Curve Sketching | mathhints.com

Extreme Value Theorem, Rolle’s Theorem, and Mean Value Theorem (other section)	Concavity and the Second Derivative
Introduction to Curve Sketching	Curve Sketching: General Rules
Relative Extrema and First Derivative Test

Introduction to Curve Sketching

Again, Curve Sketching in Calculus makes us “appreciate the math” and helps visualize what is going on with function and their derivatives. It’s not my favorite part of calculus, but it’s a necessary evil :).

Before we get started, here are some basic graphs to look at. The graphs below show example original functions, and their derivatives (slopes). Notice how when we take the derivative, we go from a quadratic (2^nd degree) to a line (1^st degree), and from a cubic (3^rd degree) to a quadratic (2^nd degree). This makes sense, since the we are always going down a degree when we take a derivative. Also notice what happens where the derivative doesn’t exist (the last graph, an absolute value function).

Relative Extrema and the First Derivative Test

Extrema and Critical Numbers

We first talked about Extrema of Functions in the Advanced Functions section here. Extrema is just a fancy word for finding the lowest (minimum) or highest (maximum) $ \boldsymbol {y}$-value in a function or interval of a function; we will use Extrema to find Critical Numbers.

We can talk about absolute extrema, or relative (local) extrema. Think of the absolute extrema as the absolute lowest or highest point in the whole domain of the function, and the relative (local) extrema as the lowest or highest for a part of the graph. Technically, relative extrema must be the minimum or maximum of a point from both sides of $ \boldsymbol {x}$, so they can’t be endpoints; they are just “valleys” or “hills”. Note that not every function has a lowest (minimum) or highest (maximum) point in an interval or even the whole domain (like the function $ y=x$), so there may not be any extrema. The endpoints of a function may be the lowest or highest points (thus the absolute, not relative extrema); these are called the endpoint extrema.

When we find the minimum and maximum values in an interval, we can use this information to find where a function is decreasing and increasing, since at a minimum or maximum value, the function takes a turn from “down to up” or “up to down”.

Here is a graph that shows some examples of absolute/relative extrema; note also the endpoint extrema points:

Critical numbers or critical points exist where a function has a minimum or maximum, whether or not the function is differentiable at that point. And it turns out that if a function is differentiable at a certain point, and that point is a minimum or maximum, the derivative at that point is 0. Here is the formal definition of a critical number:

Let $ f$ be defined at point $ c$. If $ {f}’\left( c \right)=0$, or if $ f$ is not differentiable at point $ c$, then $ c$ is a critical number of $ f$.

Increasing and Decreasing Functions, and the First Derivative Test

We talked about critical points (critical numbers) of a function (minimums or maximums), where the first derivative is 0 (or not defined). Now let’s talk about the derivative when the function is increasing (going upward from left to right), or decreasing (going downward from left to right).

When a function is increasing, the derivative (slope) is positive. When a function is decreasing, the derivative (slope) is negative. When a function is constant, or staying the same, the derivative (slope) is 0.

Here are the guidelines for finding intervals for which a function is increasing or decreasing: For a function $ f$ that is continuous on interval $ [a,b]$ and differentiable on interval $ (a,b)$, to find the intervals for which $ f$ is increasing or decreasing:

Find the critical points (minimums or maximums) in $ (a,b)$, and use these numbers to find test intervals.
For each of these test intervals, find the sign of the derivative at one test value.
If $ {f}’\left( x \right)>0$, then $ f$ is increasing on $ [a,b]$, if $ {f}’\left( x \right)<0$, then $ f$ is decreasing on $ [a,b]$, and if $ {f}’\left( x \right)=0$, then $ f$ is constant on $ [a,b]$.

Based on these guidelines, here is the First Derivative Test, which allows us to find relative minimums and maximums (also known as local minimums and local maximums).

First Derivative Test

Assume that $ c$ is a critical number of a function that is continuous on an open interval, and $ f$ is differentiable on the interval, except possibly at $ c$.

If $ {f}’\left( x \right)$ changes from negative to positive at critical point $ x=c$, then $ f$ has a relative minimum at $ x=c$.
If $ {f}’\left( x \right)$ changes from positive to negative at critical point $ x=c$, then $ f$ has a relative maximum at $ x=c$.
If $ {f}’\left( x \right)$ is positive on both sides of $ x=c$, or negative on both sides of $ x=c$, then that point is neither a relative minimum or relative maximum.

Let’s think about why this makes sense. If we have a point where a function goes from falling to rising (negative to positive slope), that point would be a minimum. Similarly, if we have a point where a function goes from rising to falling (positive to negative slope), that point would be a maximum:

Here are some problems; notice that we are using sign charts to determine the intervals that the function is decreasing or increasing. Sometimes we need to use values that aren’t even in the domain of the function (like in vertical asymptotes) in the sign charts; theoretically, these aren’t critical numbers, since they don’t exist in the original function.

First Derivative Test Problem and Graph	Solution
Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing. Use the First Derivative Test to identify all relative extrema. $ f\left( x \right)={{x}^{2}}-8x$	First take the derivative of the function to find possible critical values: $ f\left( x \right)={{x}^{2}}-8x;\,\,{f}’\left( x \right)=2x-8$. $ 2x-8=0;\,\,x=4$; thus, 4 is a critical value. Create sign chart by testing random points in intervals, using the derivative. For example, test the value “0” to see that $ 2\left( 0 \right)-8$ is negative, and test the value “5” to see that $ 2\left( 5 \right)-8$ is positive: The function is decreasing in the interval $ \left( {-\infty ,4} \right)$ and increasing in the interval $ \left( {4,\infty } \right)$. From the First Derivative Test, since we are going from a negative slope to a positive slope, $ \left( {4,{{4}^{2}}-8\cdot 4} \right)=\left( {4,-16} \right)$ is a relative extremum and is a minimum.
Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing. Use the First Derivative Test to identify all relative extrema. $ \displaystyle f\left( x \right)=\frac{{{{x}^{2}}}}{{{{x}^{2}}-4}}$	Take the derivative: $ \displaystyle f\left( x \right)=\frac{{{{x}^{2}}}}{{{{x}^{2}}-4}};\,\,{f}’\left( x \right)=\frac{{\left( {{{x}^{2}}-4} \right)\cdot 2x-{{x}^{2}}\cdot 2x}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{2}}}}=-\frac{{8x}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{2}}}}$. Find critical numbers: $ \displaystyle -\frac{{8x}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{2}}}}=0;\,\,\,-8x=0;\,\,x=0$ (Notice that at $ x=\pm \,2$, the derivative is undefined, and the original function is discontinuous, since these are asymptotes, but we’ll still use these in the sign chart. These are not critical numbers.) Create sign chart with both critical value and discontinuity values and check intervals, using derivative: Thus, the function is increasing in the intervals $ \left( {-\infty ,-2} \right)$ and $ \left( {-2,0} \right)$, and decreasing in the intervals $ \left( {0,2} \right)$ and $ \left( {2,\infty } \right)$. From the First Derivative Test, since we are going from a positive slope to a negative slope, $ \left( {0,0} \right)$ is a relative extremum and is a maximum.
Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing. Use the First Derivative Test to identify all relative extrema. $ f\left( x \right)={{x}^{{\frac{1}{3}}}}+3$	Take the derivative: $ \displaystyle f\left( x \right)={{x}^{{\frac{1}{3}}}}+3;\,\,\,\,{f}’\left( x \right)=\frac{1}{3}{{x}^{{-\frac{2}{3}}}}$. Find critical numbers: $ \displaystyle \frac{1}{3}{{x}^{{-\frac{2}{3}}}}=0;\,\,\,{{x}^{{-\frac{2}{3}}}}=0;\,\,\,\frac{1}{{\sqrt[3]{{{{x}^{2}}}}}}=0;\,\,x\text{ is undefined}$. The derivative is undefined at $ x=0$, so this is a critical number. Create sign chart with critical value and check intervals, using derivative: Thus, the function is increasing in the interval $ \left( {-\infty ,\infty } \right)$, since the original function is defined at $ x=0$. From the First Derivative Test, since both intervals have a positive slope, there are no relative extrema.

First Derivative Test Problem and Graph

Solution

Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing. Use the First Derivative Test to identify all relative extrema.

$ f\left( x \right)={{x}^{2}}-8x$

First take the derivative of the function to find possible critical values: $ f\left( x \right)={{x}^{2}}-8x;\,\,{f}’\left( x \right)=2x-8$. $ 2x-8=0;\,\,x=4$; thus, 4 is a critical value.

Create sign chart by testing random points in intervals, using the derivative. For example, test the value “0” to see that $ 2\left( 0 \right)-8$ is negative, and test the value “5” to see that $ 2\left( 5 \right)-8$ is positive:

The function is decreasing in the interval $ \left( {-\infty ,4} \right)$ and increasing in the interval $ \left( {4,\infty } \right)$. From the First Derivative Test, since we are going from a negative slope to a positive slope, $ \left( {4,{{4}^{2}}-8\cdot 4} \right)=\left( {4,-16} \right)$ is a relative extremum and is a minimum.

Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing. Use the First Derivative Test to identify all relative extrema.

$ \displaystyle f\left( x \right)=\frac{{{{x}^{2}}}}{{{{x}^{2}}-4}}$

Take the derivative: $ \displaystyle f\left( x \right)=\frac{{{{x}^{2}}}}{{{{x}^{2}}-4}};\,\,{f}’\left( x \right)=\frac{{\left( {{{x}^{2}}-4} \right)\cdot 2x-{{x}^{2}}\cdot 2x}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{2}}}}=-\frac{{8x}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{2}}}}$. Find critical numbers: $ \displaystyle -\frac{{8x}}{{{{{\left( {{{x}^{2}}-4} \right)}}^{2}}}}=0;\,\,\,-8x=0;\,\,x=0$ (Notice that at $ x=\pm \,2$, the derivative is undefined, and the original function is discontinuous, since these are asymptotes, but we’ll still use these in the sign chart. These are not critical numbers.)

Create sign chart with both critical value and discontinuity values and check intervals, using derivative:

Thus, the function is increasing in the intervals $ \left( {-\infty ,-2} \right)$ and $ \left( {-2,0} \right)$, and decreasing in the intervals $ \left( {0,2} \right)$ and $ \left( {2,\infty } \right)$. From the First Derivative Test, since we are going from a positive slope to a negative slope, $ \left( {0,0} \right)$ is a relative extremum and is a maximum.

Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing. Use the First Derivative Test to identify all relative extrema.

$ f\left( x \right)={{x}^{{\frac{1}{3}}}}+3$

Take the derivative: $ \displaystyle f\left( x \right)={{x}^{{\frac{1}{3}}}}+3;\,\,\,\,{f}’\left( x \right)=\frac{1}{3}{{x}^{{-\frac{2}{3}}}}$. Find critical numbers: $ \displaystyle \frac{1}{3}{{x}^{{-\frac{2}{3}}}}=0;\,\,\,{{x}^{{-\frac{2}{3}}}}=0;\,\,\,\frac{1}{{\sqrt[3]{{{{x}^{2}}}}}}=0;\,\,x\text{ is undefined}$. The derivative is undefined at $ x=0$, so this is a critical number.

Create sign chart with critical value and check intervals, using derivative:

Thus, the function is increasing in the interval $ \left( {-\infty ,\infty } \right)$, since the original function is defined at $ x=0$. From the First Derivative Test, since both intervals have a positive slope, there are no relative extrema.

Here’s a First Derivative Problem with a trigonometric function:

First Derivative Trig Problem and Graph	Solution
Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing in the interval $ \left( {0,2\pi } \right)$. Use the First Derivative Test to identify all relative extrema. $ \displaystyle f\left( x \right)=\frac{x}{2}+\sin x$	First take the derivative of the function to find possible critical values: $ \displaystyle f\left( x \right)=\frac{x}{2}+\sin x;\,\,\,{f}’\left( x \right)=\frac{1}{2}+\cos x$. $ \displaystyle \cos x=-\frac{1}{2};\,\,\,x=\frac{{2\pi }}{3},\,\,\frac{{4\pi }}{3}$, so these are critical values. Create sign chart by testing intervals, in interval $ \left( {0,2\pi } \right)$, using the derivative: The function is increasing in the intervals $ \displaystyle \left( {0,\frac{{2\pi }}{3}} \right)$ and $ \displaystyle \left( {\frac{{4\pi }}{3},2\pi } \right)$, and decreasing in the interval $ \displaystyle \left( {\frac{{2\pi }}{3},\frac{{4\pi }}{3}} \right)$. From the First Derivative Test, since we are going from a positive slope to negative slope at $ \displaystyle x=\frac{{2\pi }}{3},\,\,\left[ {\frac{{2\pi }}{3},\frac{{\displaystyle \frac{{2\pi }}{3}}}{2}+\sin \left( {\frac{{2\pi }}{3}} \right)} \right]\,=\left( {\frac{{2\pi }}{3},\frac{\pi }{3}+\frac{{\sqrt{3}}}{2}} \right)$ is a relative maximum. Since we are going from a negative slope to a positive slope at $ \displaystyle x=\frac{{4\pi }}{3},\,\,\left[ {\frac{{4\pi }}{3},\frac{{\displaystyle \frac{{4\pi }}{3}}}{2}+\sin \left( {\frac{{4\pi }}{3}} \right)} \right]=\left( {\frac{{4\pi }}{3},\frac{{2\pi }}{3}-\frac{{\sqrt{3}}}{2}} \right)$ is a relative minimum.

First Derivative Trig Problem and Graph

Solution

Find the critical numbers, if any, and find the open interval(s) where the function is decreasing or increasing in the interval $ \left( {0,2\pi } \right)$.

Use the First Derivative Test to identify all relative extrema.

$ \displaystyle f\left( x \right)=\frac{x}{2}+\sin x$

First take the derivative of the function to find possible critical values: $ \displaystyle f\left( x \right)=\frac{x}{2}+\sin x;\,\,\,{f}’\left( x \right)=\frac{1}{2}+\cos x$. $ \displaystyle \cos x=-\frac{1}{2};\,\,\,x=\frac{{2\pi }}{3},\,\,\frac{{4\pi }}{3}$, so these are critical values.

Create sign chart by testing intervals, in interval $ \left( {0,2\pi } \right)$, using the derivative:

The function is increasing in the intervals $ \displaystyle \left( {0,\frac{{2\pi }}{3}} \right)$ and $ \displaystyle \left( {\frac{{4\pi }}{3},2\pi } \right)$, and decreasing in the interval $ \displaystyle \left( {\frac{{2\pi }}{3},\frac{{4\pi }}{3}} \right)$. From the First Derivative Test, since we are going from a positive slope to negative slope at $ \displaystyle x=\frac{{2\pi }}{3},\,\,\left[ {\frac{{2\pi }}{3},\frac{{\displaystyle \frac{{2\pi }}{3}}}{2}+\sin \left( {\frac{{2\pi }}{3}} \right)} \right]\,=\left( {\frac{{2\pi }}{3},\frac{\pi }{3}+\frac{{\sqrt{3}}}{2}} \right)$ is a relative maximum. Since we are going from a negative slope to a positive slope at $ \displaystyle x=\frac{{4\pi }}{3},\,\,\left[ {\frac{{4\pi }}{3},\frac{{\displaystyle \frac{{4\pi }}{3}}}{2}+\sin \left( {\frac{{4\pi }}{3}} \right)} \right]=\left( {\frac{{4\pi }}{3},\frac{{2\pi }}{3}-\frac{{\sqrt{3}}}{2}} \right)$ is a relative minimum.

Concavity and the Second Derivative

I like to think of concavity as “cup up” or “cup down”. Think of concave upwards of a cup that can hold water at all points, and concave downward is a cup that empties water out at all point. It turns out that when a graph is concave upward (cup up), its slope (first derivative) is increasing, so its second derivative is positive. When a graph is concave downward (cup down), its slope is decreasing, so its second derivative is negative.

A point of inflection (POI) is exactly where the concavity changes from concave up to concave down or concave down to concave up. It turns out that a graph crosses its tangent line at a POI.

Here’s an illustration of concavity:

The Second Derivative Test can also be used in curve sketching to find relative minima and relative maxima, and is the following:

Second Derivative Test

Let $ f$ be a function whose second derivative exists on an open interval that contains $ c$:

If $ {{f}^{\prime \prime}(c)}>0$, $ f$ has a relative minimum at $ x=c$. (Think “cup up”)
If $ {{f}^{\prime \prime}(c)}<0$, $ f$ has a relative maximum at $ x=c$. (Think “cup down”)

Note that the second derivative test does not necessarily work when the first and second derivatives are 0 or undefined; the concavity of the function must change at a point of inflection. It also doesn’t say what happens at the endpoints of a function.

You might see problems like this on concavity, points of inflection, and the Second Derivative Test. I think the best way to tackle these problems is to create a sign chart using points where the first derivative is 0 or undefined (critical values) and also where the second derivative is 0 or undefined:

Second Derivative Problem and Graph	Solution
Find the point of inflection and discuss the concavity of the function: $ f\left( x \right)={{x}^{3}}-5{{x}^{2}}+8x$	$ \begin{align}f\left( x \right)&={{x}^{3}}-5{{x}^{2}}+8x\\{f}’\left( x \right)&=3{{x}^{2}}-10x+8;\,\text{Critical values:}\,\,\left( {3x-4} \right)\left( {x-2} \right)=0;\,\,x=\frac{4}{3},\,\,2\\{{f}^{\prime \prime}(x)}&=6x-10;\,\text{Point of Inflection (POI):}\,\,\,6x-10=0;\,\,x=\frac{5}{3}\end{align}$ Create sign chart by testing intervals for both first and second derivatives: The point of inflection, where the graph changes from “cup down” (concave down) to “cup up” (concave up) is at $ \displaystyle x=\frac{5}{3}$. The graph is concave down in the interval $ \displaystyle \left( {-\infty ,\frac{5}{3}} \right)$ and is concave up in the interval $ \displaystyle \left( {\frac{5}{3},\infty } \right)$. The graph is increasing in the intervals $ \displaystyle \left( {-\infty ,\frac{4}{3}} \right)$ and $ \left( {2,\infty } \right)$ and decreasing in the interval $ \displaystyle \left( {\frac{4}{3},2} \right)$. Note that $ \displaystyle \left( {\frac{4}{3},{{{\left[ {\frac{4}{3}} \right]}}^{3}}-5\cdot {{{\left[ {\frac{4}{3}} \right]}}^{2}}+8\cdot \frac{4}{3}} \right)=\left( {\frac{4}{3},\frac{{112}}{{27}}} \right)$ is a relative maximum and $ \left( {2,\,{{2}^{3}}-5\cdot {{2}^{2}}+8\cdot 2} \right)=\left( {2,4} \right)$ is a relative minimum.
Find the point of inflection and discuss the concavity of the function: $ \displaystyle f\left( x \right)=\frac{2}{{{{x}^{2}}+1}}$	$ \displaystyle \begin{align}f\left( x \right)&=\frac{2}{{{{x}^{2}}+1}}=2{{\left( {{{x}^{2}}+1} \right)}^{{-1}}}\\{f}’\left( x \right)&=-2{{\left( {{{x}^{2}}+1} \right)}^{{-2}}}\cdot 2x=-4x{{\left( {{{x}^{2}}+1} \right)}^{{-2}}}=\frac{{-4x}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{2}}}};\,\,\text{Critical value :}\,\,x=0\\{{f}^{\prime \prime}(x)}&=\frac{{{{{\left( {{{x}^{2}}+1} \right)}}^{2}}\left( {-4} \right)-\left( {-4x} \right)\left[ {2\left( {{{x}^{2}}+1} \right)\left( {2x} \right)} \right]}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{4}}}}=\frac{{4\left( {3{{x}^{2}}-1} \right)}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{3}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\text{POI: }\,\,3{{x}^{2}}-1=0;\,\,x=\pm \frac{{\sqrt{3}}}{3}\text{ }\end{align}$ By checking points in intervals, the graph is concave up in the intervals $ \displaystyle \left( {-\infty ,-\frac{{\sqrt{3}}}{3}} \right)$ and $ \displaystyle \left( {\frac{{\sqrt{3}}}{3},\infty } \right)$, and is concave down in the interval $ \displaystyle \left( {-\frac{{\sqrt{3}}}{3},\frac{{\sqrt{3}}}{3}} \right)$, with a relative maximum at $ \left( {0,2} \right)$.

Second Derivative Problem and Graph

Solution

Find the point of inflection and discuss the concavity of the function:

$ f\left( x \right)={{x}^{3}}-5{{x}^{2}}+8x$

$ \begin{align}f\left( x \right)&={{x}^{3}}-5{{x}^{2}}+8x\\{f}’\left( x \right)&=3{{x}^{2}}-10x+8;\,\text{Critical values:}\,\,\left( {3x-4} \right)\left( {x-2} \right)=0;\,\,x=\frac{4}{3},\,\,2\\{{f}^{\prime \prime}(x)}&=6x-10;\,\text{Point of Inflection (POI):}\,\,\,6x-10=0;\,\,x=\frac{5}{3}\end{align}$

Create sign chart by testing intervals for both first and second derivatives:

The point of inflection, where the graph changes from “cup down” (concave down) to “cup up” (concave up) is at $ \displaystyle x=\frac{5}{3}$. The graph is concave down in the interval $ \displaystyle \left( {-\infty ,\frac{5}{3}} \right)$ and is concave up in the interval $ \displaystyle \left( {\frac{5}{3},\infty } \right)$. The graph is increasing in the intervals $ \displaystyle \left( {-\infty ,\frac{4}{3}} \right)$ and $ \left( {2,\infty } \right)$ and decreasing in the interval $ \displaystyle \left( {\frac{4}{3},2} \right)$. Note that $ \displaystyle \left( {\frac{4}{3},{{{\left[ {\frac{4}{3}} \right]}}^{3}}-5\cdot {{{\left[ {\frac{4}{3}} \right]}}^{2}}+8\cdot \frac{4}{3}} \right)=\left( {\frac{4}{3},\frac{{112}}{{27}}} \right)$ is a relative maximum and $ \left( {2,\,{{2}^{3}}-5\cdot {{2}^{2}}+8\cdot 2} \right)=\left( {2,4} \right)$ is a relative minimum.

Find the point of inflection and discuss the concavity of the function:

$ \displaystyle f\left( x \right)=\frac{2}{{{{x}^{2}}+1}}$

$ \displaystyle \begin{align}f\left( x \right)&=\frac{2}{{{{x}^{2}}+1}}=2{{\left( {{{x}^{2}}+1} \right)}^{{-1}}}\\{f}’\left( x \right)&=-2{{\left( {{{x}^{2}}+1} \right)}^{{-2}}}\cdot 2x=-4x{{\left( {{{x}^{2}}+1} \right)}^{{-2}}}=\frac{{-4x}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{2}}}};\,\,\text{Critical value :}\,\,x=0\\{{f}^{\prime \prime}(x)}&=\frac{{{{{\left( {{{x}^{2}}+1} \right)}}^{2}}\left( {-4} \right)-\left( {-4x} \right)\left[ {2\left( {{{x}^{2}}+1} \right)\left( {2x} \right)} \right]}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{4}}}}=\frac{{4\left( {3{{x}^{2}}-1} \right)}}{{{{{\left( {{{x}^{2}}+1} \right)}}^{3}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\text{POI: }\,\,3{{x}^{2}}-1=0;\,\,x=\pm \frac{{\sqrt{3}}}{3}\text{ }\end{align}$

By checking points in intervals, the graph is concave up in the intervals $ \displaystyle \left( {-\infty ,-\frac{{\sqrt{3}}}{3}} \right)$ and $ \displaystyle \left( {\frac{{\sqrt{3}}}{3},\infty } \right)$, and is concave down in the interval $ \displaystyle \left( {-\frac{{\sqrt{3}}}{3},\frac{{\sqrt{3}}}{3}} \right)$, with a relative maximum at $ \left( {0,2} \right)$.

Here’s a concavity problem with a trigonometric function:

Trig Second Derivative Problem and Graph	Solution
Find the point of inflection and discuss the concavity of the function: $ \displaystyle f\left( x \right)=\cos \left( {\frac{x}{2}} \right);\,\,\,\,\,\left[ {0,4\pi } \right]$	$ \require {cancel} \begin{align}f\left( x \right)&=\cos \left( {\frac{x}{2}} \right);\,\,\,\,0\le x\le 4\pi \\{f}’\left( x \right)&=-\frac{1}{2}\sin \left( {\frac{x}{2}} \right)\\\text{Critical values:}&\,\,\,-\frac{1}{2}\sin \left( {\frac{x}{2}} \right)=0;\,\,\,\,\frac{x}{2}=0,\,\pi ,2\pi ,3\pi ,4\pi \\\,x&=0,2\pi ,4\pi ,\cancel{{6\pi }},\cancel{{8\pi }}\end{align}$ $ \begin{align}{{f}^{\prime \prime}(x)}&=-\frac{1}{4}\cos \left( {\frac{x}{2}} \right)\,\\\text{Point of Inflection (POI):}&\,\,\,-\frac{1}{4}\cos \left( {\frac{x}{2}} \right)=0;\,\,\,\,\frac{x}{2}=\frac{\pi }{2},\frac{{3\pi }}{2},\frac{{5\pi }}{2},\frac{{7\pi }}{2}\\\,\,\,\,x&=\pi ,3\pi ,\cancel{{5\pi }},\cancel{{7\pi }}\end{align}$ Create sign chart by testing intervals for both first and second derivatives: The points of inflection, where the graph changes from “cup down” (concave down) to “cup up” (concave up) are at $ x=\pi ,\,3\pi $. The graph is concave down in the intervals $ \left( {0,\pi } \right)$ and $ \left( {3\pi ,4\pi } \right)$, and is concave up in the interval $ \left( {\pi ,3\pi } \right)$. The graph is decreasing in the interval $ \left( {0,2\pi } \right)$, and increasing in the interval $ \left( {2\pi ,4\pi } \right)$. Note that $ \displaystyle \left( {0,\cos \left( {\frac{0}{2}} \right)} \right)=\left( {0,1} \right)$ and $ \displaystyle \left( {4\pi ,\cos \left( {\frac{{4\pi }}{2}} \right)} \right)=\left( {4\pi ,1} \right)$ are relative maximums (endpoints) and $ \displaystyle \left( {2\pi ,\cos \left( {\frac{{2\pi }}{2}} \right)} \right)=\left( {2\pi ,-1} \right)$ is a relative minimum.

Trig Second Derivative Problem and Graph

Solution

Find the point of inflection and discuss the concavity of the function:

$ \displaystyle f\left( x \right)=\cos \left( {\frac{x}{2}} \right);\,\,\,\,\,\left[ {0,4\pi } \right]$

$ \require {cancel} \begin{align}f\left( x \right)&=\cos \left( {\frac{x}{2}} \right);\,\,\,\,0\le x\le 4\pi \\{f}’\left( x \right)&=-\frac{1}{2}\sin \left( {\frac{x}{2}} \right)\\\text{Critical values:}&\,\,\,-\frac{1}{2}\sin \left( {\frac{x}{2}} \right)=0;\,\,\,\,\frac{x}{2}=0,\,\pi ,2\pi ,3\pi ,4\pi \\\,x&=0,2\pi ,4\pi ,\cancel{{6\pi }},\cancel{{8\pi }}\end{align}$

$ \begin{align}{{f}^{\prime \prime}(x)}&=-\frac{1}{4}\cos \left( {\frac{x}{2}} \right)\,\\\text{Point of Inflection (POI):}&\,\,\,-\frac{1}{4}\cos \left( {\frac{x}{2}} \right)=0;\,\,\,\,\frac{x}{2}=\frac{\pi }{2},\frac{{3\pi }}{2},\frac{{5\pi }}{2},\frac{{7\pi }}{2}\\\,\,\,\,x&=\pi ,3\pi ,\cancel{{5\pi }},\cancel{{7\pi }}\end{align}$

Create sign chart by testing intervals for both first and second derivatives:

The points of inflection, where the graph changes from “cup down” (concave down) to “cup up” (concave up) are at $ x=\pi ,\,3\pi $. The graph is concave down in the intervals $ \left( {0,\pi } \right)$ and $ \left( {3\pi ,4\pi } \right)$, and is concave up in the interval $ \left( {\pi ,3\pi } \right)$. The graph is decreasing in the interval $ \left( {0,2\pi } \right)$, and increasing in the interval $ \left( {2\pi ,4\pi } \right)$. Note that $ \displaystyle \left( {0,\cos \left( {\frac{0}{2}} \right)} \right)=\left( {0,1} \right)$ and $ \displaystyle \left( {4\pi ,\cos \left( {\frac{{4\pi }}{2}} \right)} \right)=\left( {4\pi ,1} \right)$ are relative maximums (endpoints) and $ \displaystyle \left( {2\pi ,\cos \left( {\frac{{2\pi }}{2}} \right)} \right)=\left( {2\pi ,-1} \right)$ is a relative minimum.

Here are more Second Derivative Test problems:

Second Derivative Problem and Graph	Solution
Using the Second Derivative Test, find relative extrema for the following function, if it’s possible to do so: $ \displaystyle f\left( x \right)=4x+\frac{1}{x}$	$ \begin{align}f\left( x \right)&=4x+\frac{1}{x};\,\,\,f\left( x \right)=4x+{{x}^{{-1}}}\\{f}’\left( x \right)&=4-{{x}^{{-2}}};\,\,\,\,\,\text{Critical values:}\,\,\,4-\frac{1}{{{{x}^{2}}}}=0;\,\,\,\,x=\pm \displaystyle \frac{1}{2}\\{{f}^{\prime \prime}(x)}&=2{{x}^{{-3}}};\,\,\,\,\,\,\,\text{Plug in}\,\,\frac{1}{2}:2{{\left( {\displaystyle \frac{1}{2}} \right)}^{{-3}}}=16\,\,\text{(positive) (concave up)}\end{align}$ $ \displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Plug in}\,\,-\frac{1}{2}:2{{\left( {-\frac{1}{2}} \right)}^{{-3}}}=-16\,\,\text{(negative) (concave down)}$ $ \displaystyle \left( {\frac{1}{2},4\left( {\frac{1}{2}} \right)+\frac{1}{{\displaystyle \frac{1}{2}}}} \right)=\left( {\frac{1}{2},4} \right)$ is a relative minimum, and $ \displaystyle \left( {-\frac{1}{2},4\left( {-\displaystyle \frac{1}{2}} \right)+\frac{1}{{-\displaystyle \frac{1}{2}}}} \right)=\left( {-\frac{1}{2},-4} \right)$ is a relative maximum.
Using the Second Derivative Test, find relative extrema for the following function, if it’s possible to do so: $ \displaystyle f\left( x \right)=\frac{x}{{\sqrt{{1-{{x}^{2}}}}}}$	$ \begin{align}f\left( x \right)&=\frac{x}{{\sqrt{{1-{{x}^{2}}}}}};\,\,\,f\left( x \right)=x{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{1}{2}}}}\,\,\,\,\,\\{f}’\left( x \right)&=x\cdot -\frac{1}{2}{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{3}{2}}}}\cdot -2x+{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{1}{2}}}}\cdot 1\\&={{x}^{2}}{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{3}{2}}}}+{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{1}{2}}}}\\\,\,\,\,\,\,\,\,\,\,&={{\left( {1-{{x}^{2}}} \right)}^{{-\frac{3}{2}}}}\left( {\cancel{{{{x}^{2}}}}+1\cancel{{-{{x}^{2}}}}} \right);\,\,\,\,\,\,\,\frac{1}{{{{{\left( {1-{{x}^{2}}} \right)}}^{{\frac{3}{2}}}}}}=0\end{align}$ $ \displaystyle \text{Critical points (undefined):}\,\,\pm 1$ Note that the Second Derivative Test does not apply, since the first derivative fails to exist (undefined) at $ x=\pm 1$ (there are actually vertical asymptotes at those values). Therefore, we don’t have to take the second derivative. The first derivative will always be positive, so there are no relative extrema (maximums or minimums).

Second Derivative Problem and Graph

Solution

Using the Second Derivative Test, find relative extrema for the following function, if it’s possible to do so:

$ \displaystyle f\left( x \right)=4x+\frac{1}{x}$

$ \begin{align}f\left( x \right)&=4x+\frac{1}{x};\,\,\,f\left( x \right)=4x+{{x}^{{-1}}}\\{f}’\left( x \right)&=4-{{x}^{{-2}}};\,\,\,\,\,\text{Critical values:}\,\,\,4-\frac{1}{{{{x}^{2}}}}=0;\,\,\,\,x=\pm \displaystyle \frac{1}{2}\\{{f}^{\prime \prime}(x)}&=2{{x}^{{-3}}};\,\,\,\,\,\,\,\text{Plug in}\,\,\frac{1}{2}:2{{\left( {\displaystyle \frac{1}{2}} \right)}^{{-3}}}=16\,\,\text{(positive) (concave up)}\end{align}$

$ \displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Plug in}\,\,-\frac{1}{2}:2{{\left( {-\frac{1}{2}} \right)}^{{-3}}}=-16\,\,\text{(negative) (concave down)}$

$ \displaystyle \left( {\frac{1}{2},4\left( {\frac{1}{2}} \right)+\frac{1}{{\displaystyle \frac{1}{2}}}} \right)=\left( {\frac{1}{2},4} \right)$ is a relative minimum, and $ \displaystyle \left( {-\frac{1}{2},4\left( {-\displaystyle \frac{1}{2}} \right)+\frac{1}{{-\displaystyle \frac{1}{2}}}} \right)=\left( {-\frac{1}{2},-4} \right)$ is a relative maximum.

Using the Second Derivative Test, find relative extrema for the following function, if it’s possible to do so:

$ \displaystyle f\left( x \right)=\frac{x}{{\sqrt{{1-{{x}^{2}}}}}}$

$ \begin{align}f\left( x \right)&=\frac{x}{{\sqrt{{1-{{x}^{2}}}}}};\,\,\,f\left( x \right)=x{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{1}{2}}}}\,\,\,\,\,\\{f}’\left( x \right)&=x\cdot -\frac{1}{2}{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{3}{2}}}}\cdot -2x+{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{1}{2}}}}\cdot 1\\&={{x}^{2}}{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{3}{2}}}}+{{\left( {1-{{x}^{2}}} \right)}^{{-\frac{1}{2}}}}\\\,\,\,\,\,\,\,\,\,\,&={{\left( {1-{{x}^{2}}} \right)}^{{-\frac{3}{2}}}}\left( {\cancel{{{{x}^{2}}}}+1\cancel{{-{{x}^{2}}}}} \right);\,\,\,\,\,\,\,\frac{1}{{{{{\left( {1-{{x}^{2}}} \right)}}^{{\frac{3}{2}}}}}}=0\end{align}$

$ \displaystyle \text{Critical points (undefined):}\,\,\pm 1$

Note that the Second Derivative Test does not apply, since the first derivative fails to exist (undefined) at $ x=\pm 1$ (there are actually vertical asymptotes at those values). Therefore, we don’t have to take the second derivative. The first derivative will always be positive, so there are no relative extrema (maximums or minimums).

Here’s one more problem where the Second Derivative Test does not apply at certain points:

Second Derivative Problem	Solution
The first derivative of the function $ f$ is given by $ f\left( x \right)={{x}^{4}}+4{{x}^{3}}$. What are the $ x$-coordinates of the points of inflections of the graph of $ f$?	First, use the First Derivative Test to get the critical values (min and/or max) of the function. Since we already have the first derivative, set it to 0 and solve for $ x$: $ \begin{array}{c}{f}’\left( x \right)={{x}^{4}}+4{{x}^{3}}\\\,\text{Critical values:}\,\,\,{{x}^{4}}+4{{x}^{3}}=0;\,\,\,\,{{x}^{3}}\left( {x+4} \right)=0\\\,\,\,\,\,\,\,\,\,\,x=0,-4\end{array}$ Now, use the Second Derivative (the derivative of the first derivative), and set to 0 to get possible points of inflection: $ \require {cancel} \begin{array}{c}{{f}^{\prime \prime}(x)}=4{{x}^{3}}+12{{x}^{2}}\\\text{Point of Inflections (POI):}\,\,\,4{{x}^{3}}+12{{x}^{2}}=0;\,\,\,\,4{{x}^{2}}\left( {x+3} \right)=0\\\,\,\,\,\,\,\,\,\,\,x=\cancel{0},-3\end{array}$ Note that the Second Derivative Test does not apply at $ x=0$, since the concavity at that point doesn’t change. Therefore, the only point of inflection is at $ x=-3$. Here is a graph of what $ f$ may look like:

Second Derivative Problem

Solution

The first derivative of the function $ f$ is given by $ f\left( x \right)={{x}^{4}}+4{{x}^{3}}$.

What are the $ x$-coordinates of the points of inflections of the graph of $ f$?

First, use the First Derivative Test to get the critical values (min and/or max) of the function. Since we already have the first derivative, set it to 0 and solve for $ x$:

$ \begin{array}{c}{f}’\left( x \right)={{x}^{4}}+4{{x}^{3}}\\\,\text{Critical values:}\,\,\,{{x}^{4}}+4{{x}^{3}}=0;\,\,\,\,{{x}^{3}}\left( {x+4} \right)=0\\\,\,\,\,\,\,\,\,\,\,x=0,-4\end{array}$

Now, use the Second Derivative (the derivative of the first derivative), and set to 0 to get possible points of inflection:

$ \require {cancel} \begin{array}{c}{{f}^{\prime \prime}(x)}=4{{x}^{3}}+12{{x}^{2}}\\\text{Point of Inflections (POI):}\,\,\,4{{x}^{3}}+12{{x}^{2}}=0;\,\,\,\,4{{x}^{2}}\left( {x+3} \right)=0\\\,\,\,\,\,\,\,\,\,\,x=\cancel{0},-3\end{array}$

Note that the Second Derivative Test does not apply at $ x=0$, since the concavity at that point doesn’t change. Therefore, the only point of inflection is at $ x=-3$. Here is a graph of what $ f$ may look like:

Curve Sketching: General Rules

Here are some general curve sketching rules:

Find critical numbers (numbers that make the first derivative 0 or undefined).
Put the critical numbers in a sign chart to see where the first derivative is positive or negative (plug in the first derivative to get signs).
Where first derivative is positive, the function is increasing; where it’s negative, the function is decreasing (remember that you can combine two consecutive intervals only if the original function is defined for that critical number).
To get relative minimums and relative maximums, see how the derivative is changing. If it’s changing from negative to positive, it’s a minimum, and if it’s changing from positive to negative, it’s a maximum. To get the coordinates of the point at these places, plug the $ x$-value into the original function to get the $ y$-value.
Get the second derivative, and find the values where it’s either 0 or undefined. (Make sure these points are defined for both the function and first derivative).
Put those values in a sign chart to see where the second derivative is positive or negative (plug in the second derivative to get signs).
Where second derivative is positive, the graph is concave up, where the second derivative is negative, the graph is concave down. Remember that you can combine two consecutive intervals only if the original function is defined for that for the first and second derivative. A point of inflection (POI) occurs when the second derivate changes sign.
Use other points (can use a t-chart) to help graph!

Also, these tips may help:

If finding absolute extrema, find the critical numbers and endpoints, then plug into the original function to find $ y$-values. Compare all these values: the largest is the absolute maximum and the smallest is the absolute minimum.
When graphing, it might be helpful to identify any asymptotes or removable discontinuities (holes) by seeing what makes the denominator of the original function 0. Remember that a hole happens when you can cross out a factor in both the numerator and denominator (see Drawing Rational Graphs in the Graphing Rational Functions, including Asymptotes section).

Here are some other hints that may help with the relationship of $ f$, $ {f}’$ and $ {{f}^{\prime \prime}}$. Again, remember that the relationship of $ f$ to $ {f}’$ is the same as $ {f}’$ to $ {{f}^{\prime \prime}}$; similarly, the relationship of $ {f}’$ to $ f$ is the same as $ {{f}^{\prime \prime}}$ to$ {f}’$. Also, I like to use the PMS acronym (sorry 🙂 ) to “travel” back and forth among the curves of the function, derivative, and second derivative:

Function	Derivative	Second Derivative
P: Point of Inflection (POI)	M: Maximum or Minimum	S: Sign Change

If….	Then….	If….	Then…
$ f\left( x \right)$ is increasing	$ {f}’\left( x \right)$ is positive	$ {f}’\left( x \right)$ is increasing	$ {{f}^{\prime \prime}(x)}$ is positive
$ f\left( x \right)$ is decreasing	$ {f}’\left( x \right)$ is negative	$ {f}’\left( x \right)$ is decreasing	$ {{f}^{\prime \prime}(x)}$ is negative
$ f\left( x \right)$ has a relative min or max	$ {f}’\left( x \right)$ is 0 or does not exist (changes sign)	$ {f}’\left( x \right)$ has a relative min or max	$ {{f}^{\prime \prime}(x)}$ is 0 or does not exist (changes sign)
$ f\left( x \right)$ has a POI	$ {f}’\left( x \right)$ has a relative min or max	$ {f}’\left( x \right)$ has a POI	$ {{f}^{\prime \prime}(x)}$ has a relative min or max
$ f\left( x \right)$ has a POI	$ {{f}^{\prime \prime}(x)}$ changes sign	$ {f}’\left( x \right)$ has a relative min or max	$ f\left( x \right)$ has a POI, if concavity changes at that point
$ f\left( x \right)$ is concave up	$ {{f}^{\prime \prime}(x)}$ is positive	$ f\left( x \right)$ is concave down	$ {{f}^{\prime \prime}(x)}$ is negative
$ {f}’\left( x \right)$ changes from from negative to positive	$ f\left( x \right)$ has a relative min	$ {f}’\left( x \right)$ changes from from positive to negative	$ f\left( x \right)$ has a relative max
$ {{f}^{\prime \prime}(x)}$ changes from from negative to positive	$ {f}’\left( x \right)$ has a relative min	$ {{f}^{\prime \prime}(x)}$ changes from from positive to negative	$ {f}’\left( x \right)$ has a relative max
$ {f}’\left( x \right)$ is 0	horizontal tangent; $ f\left( x \right)$ could be relative min/max or a POI	$ {{f}^{\prime \prime}(x)}$ has a relative min or max	$ {f}’\left( x \right)$ has a POI, if concavity changes at that point
$ f\left( x \right)$ exists, $ {f}’\left( x \right)$ doesn’t exist	possibly a vertical tangent, or an absolute max or min	$ {f}’\left( x \right)$ is 0 and $ {{f}^{\prime \prime}(x)}$ is 0	$ {{f}^{\prime \prime}(x)}$ has a POI, if concavity changes at that point

Here’s a nice visual for curve sketching that shows the relationship of a function, derivative, and second derivative for the four curves shown:

Here are some problems that you may see:

Curve Sketching Problem	Graph	Explanation
Sketch a curve: $ \begin{array}{c}f\left( 2 \right)=-1\,\,\,\text{and }{f}’\left( 2 \right)=0\\{f}’\left( x \right)>0\,\,\,\text{for }x\ne 2\\{{f}^{\prime \prime}(x)}<0\,\,\,\text{for }x<2\text{ }\\{{f}^{\prime \prime}(x)}>0\,\,\,\text{for }x>2\text{ }\end{array}$		Since $ {f}’\left( 2 \right)=0$, but the function has a positive slope for everywhere except for $ x=2$, the point $ \left( {2,-1} \right)$ must be a point of inflections (POI). The function is “concave down” (second derivative is negative) for $ x<2$ and “concave up” (second derivative is positive) for $ x>2$.
Sketch a curve: $ \begin{array}{c}{f}’\left( x \right)>0\,\,\,\text{for }x<0\,\,\,\text{and }x>2\\{f}’\left( x \right)<0\,\,\,\text{for }0<x<2\\{{f}^{\prime \prime}(x)}>0\,\,\,\text{for }x\ne 0\\f\left( 0 \right)=4\,\,\text{but }{f}’\left( 0 \right)\text{ is undefined }\\\text{Horizontal Tangent at}\left( {2,-1} \right)\end{array}$		The function is increasing everywhere, except between 0 and 2. It is “concave up” for everywhere, except where $ x=0$. Since the derivative isn’t defined at $ x=0$, there is some sort of a sharp turn. The horizontal tangent at $ \left( {2,-1} \right)$ indicates a minimum in this case.

Curve Sketching Problem

Graph

Explanation

Sketch a curve:

$ \begin{array}{c}f\left( 2 \right)=-1\,\,\,\text{and }{f}’\left( 2 \right)=0\\{f}’\left( x \right)>0\,\,\,\text{for }x\ne 2\\{{f}^{\prime \prime}(x)}<0\,\,\,\text{for }x<2\text{ }\\{{f}^{\prime \prime}(x)}>0\,\,\,\text{for }x>2\text{ }\end{array}$

Since $ {f}’\left( 2 \right)=0$, but the function has a positive slope for everywhere except for $ x=2$, the point $ \left( {2,-1} \right)$ must be a point of inflections (POI).

The function is “concave down” (second derivative is negative) for $ x<2$ and “concave up” (second derivative is positive) for $ x>2$.

Sketch a curve:

$ \begin{array}{c}{f}’\left( x \right)>0\,\,\,\text{for }x<0\,\,\,\text{and }x>2\\{f}’\left( x \right)<0\,\,\,\text{for }0<x<2\\{{f}^{\prime \prime}(x)}>0\,\,\,\text{for }x\ne 0\\f\left( 0 \right)=4\,\,\text{but }{f}’\left( 0 \right)\text{ is undefined }\\\text{Horizontal Tangent at}\left( {2,-1} \right)\end{array}$

The function is increasing everywhere, except between 0 and 2. It is “concave up” for everywhere, except where $ x=0$.

Since the derivative isn’t defined at $ x=0$, there is some sort of a sharp turn.

The horizontal tangent at $ \left( {2,-1} \right)$ indicates a minimum in this case.

Here’s one where we might have the information in a table.

Sketch a possible graph:

	$ x<-1$	$ x=-1$	$ -1<x<2$	$ x=2$	$ 2<x<5$	$ x=5$	$ 5<x<8$	$ x=8$
$ f\left( x \right)$	negative	undefined	positive	1		–2		1
$ {f}’\left( x \right)$	negative	undefined	negative	0	negative	undefined	positive
$ {{f}^{\prime \prime}(x)}$	negative	undefined	positive	0	negative		negative

Here is what the graph might look like:

Curve Sketching Graph	Explanation
	Plot the points first, and then sketch the rising and falling intervals, based on the first derivative. Then look at the second derivative to shape the intervals, either concave up or concave down. Note that at $ x=-1$, since the function, derivative, and second derivative are all undefined, we could have either a discontinuous function (as shown), or possibly an asymptote.

Curve Sketching Graph

Explanation

Plot the points first, and then sketch the rising and falling intervals, based on the first derivative.

Then look at the second derivative to shape the intervals, either concave up or concave down.

Note that at $ x=-1$, since the function, derivative, and second derivative are all undefined, we could have either a discontinuous function (as shown), or possibly an asymptote.

Here are more types of curve sketching problems you may see:

Curve Sketching Problem:

Find all relative extrema, with $ y$-values, and points of inflection for $ f\left( x \right)=6{{x}^{{\frac{2}{3}}}}-x$. Find where the function is increasing and decreasing, and where it is concave up and down.

Solution:

First take the first and second derivatives: $ \displaystyle f\left( x \right)=6{{x}^{{\frac{2}{3}}}}-x;\,\,\,\,{{f}^{\prime}(x)}=4{{x}^{{-\frac{1}{3}}}}-1;\,\,\,\,\,{{f}^{\prime \prime}(x)}=-\frac{4}{3}{{x}^{{-\frac{4}{3}}}}$. Next, find the increasing (decreasing) intervals: where the derivative is positive (negative). Set the first derivative to 0 first to find critical point(s), also notifying that the derivative is undefined at at $ x=0$. Thus, the critical points are at $ x=0$ and $ x=64$. Use test points to find increasing/decreasing intervals.

$ \displaystyle 4{{x}^{{-\frac{1}{3}}}}-1=0;\,\,\,\,{{\left( {4{{x}^{{-\frac{1}{3}}}}} \right)}^{{-3}}}={{\left( 1 \right)}^{{-3}}};\,\,\,\displaystyle \frac{1}{{64}}x=1;\,\,\,\,x=64$: .

To get the $ y$ points, plug in $ x$ in the original equation: $ \displaystyle y=6{{\left( 0 \right)}^{{\frac{2}{3}}}}-0=0;\,\,\,\,y=6{{\left( {64} \right)}^{{\frac{2}{3}}}}-64=32$. The critical points are $ \left( {0,0} \right)$ (min) and $ \left( {64,32} \right)$ (max).

Set the second derivative to 0 to find point of inflections (POI) and intervals of concavity, using test points with the second derivative: $ \displaystyle {4 \over 3}{x^{ – {4 \over 3}}} = 0;\,\,\,\,\,{\left( {{4 \over 3}{x^{ – {4 \over 3}}}} \right)^{ – {3 \over 4}}} = {0^{ – {3 \over 4}}}$ (again, noticing that the second derivative is undefined at $ x=0$): . Since both intervals are negative, there is not a POI.

We have the following characteristics; notice how we have to “jump” over (exclude) $ x=0$ in some cases:

Increasing Intervals: $ \left( {0,64} \right)$ Decreasing Intervals: $ \left( {-\infty ,0} \right)\cup \left( {64,\infty } \right)$

Local Minimum: $ \left( {0,0} \right)$ Local Maximum: $ \left( {64,32} \right)$

Concave Up Intervals: none Concave Down Intervals: $ \left( {-\infty ,0} \right)\cup \left( {0,\infty } \right)$ Points of Inflection (POI): none

Curve Sketching Problem:

The graph $ {f}’\left( x \right)$ to the right is the derivative of a function $ f$.

a) The graph $ f$ has a local maximum at $ x=$?

b) The graph $ f$ has a local minimum at $ x=$?

c) The graph $ f$ has a point of inflection (POI) at $ x=$?

Solution:

Given the derivative, try to draw the function. Remember P→M→S: when going “backwards” from the derivative to the original function, where there are minimums/maximums in the derivative, there are POI in the original, and where there are sign changes in the derivative, there are minimums/maximums in the original.

Answer the questions from above:

a) When a function has a local maximum, its derivative has a sign change from positive to negative, so there is a local maximum at $ x=4$.

b) When a function has a local minimum, its derivative has a sign change from negative to positive, so there are no local minimums. (Note: in some textbooks, endpoints can be relative extrema, so we’d have local minimums at $ x=0$ and $ x=5$).

c) When a function has a POI, its derivative has a minimum or maximum, so there are POIs at $ x=1$ and $ x=3$.

Learn these rules, and practice, practice, practice!

On to Optimization – you are ready!