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increasing/decreasing intervals

critical points

first derivative test

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17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

Second Derivative - Practice Problems Conversion

[A01-1349] - [A02-1350] - [A03-1351] - [A04-1348] - [A05-1354] - [B01-1352] - [B02-1353]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

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concavity, inflection
\(f(x)=2x^3+6x^2-5x+1\)

Problem Statement

concavity, inflection
\(f(x)=2x^3+6x^2-5x+1\)

Solution

1349 solution video

video by Krista King Math

close solution

concavity, inflection
\(\displaystyle{f(x)=\frac{x^2+1}{x^2}}\)

Problem Statement

concavity, inflection
\(\displaystyle{f(x)=\frac{x^2+1}{x^2}}\)

Solution

1350 solution video

video by Krista King Math

close solution

concavity, inflection
\(f(x)=2+3x^2-x^3\)

Problem Statement

concavity, inflection
\(f(x)=2+3x^2-x^3\)

Solution

1351 solution video

video by PatrickJMT

close solution

concavity, inflection
\(h(x)=(x^2-1)^3\)

Problem Statement

concavity, inflection
\(h(x)=(x^2-1)^3\)

Solution

1352 solution video

video by PatrickJMT

close solution

Determine the open intervals where the function \(f(x)=x^2 e^{4x}\) is concave up or concave down and determine any points of inflection.

Problem Statement

Determine the open intervals where the function \(f(x)=x^2 e^{4x}\) is concave up or concave down and determine any points of inflection.

Solution

2290 solution video

video by MIP4U

close solution

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

Interval/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

For these problems, use the second derivative test to determine maximums and minimums, unless otherwise instructed.

maximums/minimums
\(f(x)=x^3-3x+1\)

Problem Statement

maximums/minimums
\(f(x)=x^3-3x+1\)

Solution

1348 solution video

video by Krista King Math

close solution

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 solution video

video by PatrickJMT

close solution

maximums/minimums
\(\displaystyle{f(x)=\frac{x}{x^2+4}}\)

Problem Statement

maximums/minimums
\(\displaystyle{f(x)=\frac{x}{x^2+4}}\)

Solution

1353 solution video

video by PatrickJMT

close solution
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