## 17Calculus Derivatives - Using The Second Derivative To Analyze and Graph Functions

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On this page we discuss using the second derivative to help determine what the graph of a function looks like. The techniques discussed are concavity, inflection points and the second derivative test.

Concavity

Concavity relates to how the graph is curving, either upward or downward. If a graph is curving upward, then it looks like a cup and it could hold water. The graph of $$y=x^2$$ is concave upward.

If the graph is curving downward, then it looks like an arched roof and could keep you dry you in the rain. The graph of $$y=-x^2$$ is concave downward. This plot shows several examples of concavity to help you get a feel for what they look like.

The concept of concavity parallels increasing and decreasing intervals. We learned that increasing and decreasing sections of a function change only at critical points. Similarly, concavity is related to the second derivative and changes only at inflection points.

Here is a quick video clip explaining this idea again.

### PatrickJMT - Concavity [10min-22secs]

video by PatrickJMT

Inflection Points

We use inflection points to help us determine the where concavity changes. Basically, concavity will change only at inflection points. To find inflection points, we use a similar procedure as we did for critical points, except we us the second derivative. So we start by taking the derivative twice, set the result to zero and solve for the x-values. We also look at values where the second derivative is not defined but the points are in the domain of the original function.

The inflection points allow us to determine concavity. We can use the following table format to organize the information. We assume here that we have a function $$g(x)$$ with break points at $$x=c_1$$, $$x=c_2$$ and $$x=c_3$$ and the function is defined for $$x < c_1$$ and for $$x > c_3$$.

Table Format For Inflection Points

Interval

$$-\infty < x < c_1$$

$$c_1 < x < c_2$$

$$c_2 < x < c_3$$

$$c_3 < x < \infty$$

Test x-value

Sign of $$g''(x)$$

Conclusion

Possible conclusions include concave upward or concave downward.
Note: The break points include points of inflection and discontinuities. Basically, the entire domain needs to be covered by the intervals in the first row.

The test values can be any point in the open interval in each column.

Before we go on, try these practice problems.

Practice

Unless otherwise instructed, find the points of inflection and determine concavity.

$$f(x)=2x^3+6x^2-5x+1$$

Problem Statement

Calculate the inflection points of $$f(x)=2x^3+6x^2-5x+1$$ and determine where the function is concave up or concave down.

Solution

### 1349 video

video by Krista King Math

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

Problem Statement

Calculate the inflection points of $$\displaystyle{f(x)=\frac{x^2+1}{x^2}}$$ and determine where the function is concave up or concave down.

Solution

### 1350 video

video by Krista King Math

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$$f(x)=2+3x^2-x^3$$

Problem Statement

Calculate the inflection points of $$f(x)=2+3x^2-x^3$$ and determine where the function is concave up or concave down.

Solution

### 1351 video

video by PatrickJMT

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$$h(x)=(x^2-1)^3$$

Problem Statement

Calculate the inflection points of $$h(x)=(x^2-1)^3$$ and determine where the function is concave up or concave down.

Solution

### 1352 video

video by PatrickJMT

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$$f(x)=x^2 e^{4x}$$

Problem Statement

Calculate the inflection points of $$f(x)=x^2 e^{4x}$$ and determine where the function is concave up or concave down.

Solution

### 2290 video

video by MIP4U

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Second Derivative Test

The second derivative test is interesting in that it tests for the same information as the first derivative test. On the good side it is easier to use. On the bad side, it doesn't always work.

The idea is that you find the second derivative and then plug the critical points in the second derivative. If the result is less than zero, then you have a relative maximum, greater than zero, a relative minimum. If you get zero, the test is inconclusive and you drop back and use the first derivative test. To organize your information, you can use a table in this format.

Table Format For The Second Derivative Test

Critical Point

$$x = c_1$$

$$x = c_2$$

$$x = c_3$$

Sign of $$g''(x)$$

Conclusion

The possible conclusions in the last row above maximum, minimum or inconclusive. In the last case, we need to use the first derivative test.

Before working practice problems, take a few minutes to watch this quick video clip explaining the second derivative test in more detail.

### PatrickJMT - Second Derivative Test [1min-30secs]

video by PatrickJMT

Practice

Unless otherwise instructed, use the second derivative test to determine maximums and minimums.

$$f(x)=x^3-3x+1$$

Problem Statement

Use the second derivative test to determine the extrema for $$f(x)=x^3-3x+1$$. If the second derivative test is inconclusive, use the first derivative test.

Solution

### 1348 video

video by Krista King Math

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Suppose $$f(x)$$ has a critical point at $$x=4$$ and $$f''(4) = -3$$. What can be said about $$f(x)$$ at $$x=4$$?

Problem Statement

Suppose $$f(x)$$ has a critical point at $$x=4$$ and $$f''(4) = -3$$. What can be said about $$f(x)$$ at $$x=4$$?

Solution

### 1354 video

video by PatrickJMT

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$$\displaystyle{f(x)=\frac{x}{x^2+4}}$$

Problem Statement

Use the second derivative test to determine the extrema for $$\displaystyle{f(x)=\frac{x}{x^2+4}}$$. If the second derivative test is inconclusive, use the first derivative test.

Solution

### 1353 video

video by PatrickJMT

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You CAN Ace Calculus

 derivatives basics of graphing

external links you may find helpful related topics on other pages increasing/decreasing intervals critical points first derivative test graphing youtube playlist

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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