Concavity | AP Calc AB 2nd derivative test

What you’ll learn:

How to classify relative extrema using the 2nd derivative test instead of the 1st derivative sign chart
Understanding the 2nd derivative as concavity

Once you’ve found a critical point where $f^{'} (x) = 0$ , you can often bypass a full sign chart for $f^{'} (x)$ by using the 2nd derivative test to classify extrema.

2nd derivative test for extrema

Each critical point $x = c$ can be classified by evaluating the sign of the 2nd derivative at the point, $f^{''} (c)$ :

Relative minimum: If $f^{''} (c) > 0$ , then $f (x)$ is concave up.
- Visual cue: The curve is shaped “like a cup.”
Relative maximum: If $f^{''} (c) < 0$ , then $f (x)$ is concave down.
- Visual cue: The curve is shaped “like a frown.”
The test is inconclusive if $f^{''} (x) = 0$ or is undefined.
- Use the 1st derivative test to classify the point.

Sidenote

Required condition

The 2nd derivative test only works when $f^{'} (c) = 0$ .

If $f^{'} (c)$ is undefined, then $f^{''} (c)$ is also undefined, and the test cannot be used.

As shown in the diagram below, a horizontal tangent line combined with the graph’s shape determines whether the point is a peak (relative maximum) or a valley (relative minimum).

Example 1: Classic problem

Classify the extrema of

$f (x) = \frac{1}{5} x^{5} + \frac{1}{4} x^{4}$

using the 2nd derivative test.

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = x^{4} + x^{3}$

To find the critical points, set equal to $0$ :

$x^{4} + x^{3} = 0 x^{3} (x + 1) = 0 x = 0, - 1$

Now apply the 2nd derivative test. The 2nd derivative is

$f^{''} (x) = 4 x^{3} + 3 x^{2}$

Evaluate each critical point:

$f^{''} (- 1) f^{''} (0) = - 1 = 0$

Classify each:

Since $f^{''} (- 1) < 0$ , $f (x)$ is concave down at $x = - 1$ , so it is a relative max.
Since $f^{''} (0) = 0$ , the test is inconclusive. Using the 1st derivative test,

Interval	Sign of $f^{'}$
$x < 0$	$-$
$x > 0$	$+$

Since $f^{'}$ changes from negative to positive, there is a relative min at $x = 0$ .

In this particular case, the 1st derivative test alone would have been straightforward. However, for more complicated exponential or trigonometric functions, the 2nd derivative test is often more efficient since it avoids interval testing altogether.

Example 2: Using your calculator

The derivative of a function $f$ is given by

$f^{'} (x) = 2 x^{2} - e^{c o s (x)}$

Find all critical points of $f$ and use the 2nd derivative test to classify them as relative maximums or relative minimums.

Solution

(spoiler)

Critical points of $f$ occur where $f^{'} (x) = 0$ or is undefined.

As a calculator-active question, you can simply define the given derivative of $f$ as a separate function $g$ in Desmos by typing the following:

$g (x) = 2 x^{2} - e^{c o s (x)}$

Since $g (x) = f^{'} (x)$ , note where the graph of $g$ crosses the $x$ -axis, which is where $f^{'} (x) = 0$ . These are the critical points of $x = \pm 0.947$ .

Because $g (x) = f^{'} (x)$ , then differentiating $g (x)$ gives the 2nd derivative of $f$ i.e. $g^{'} (x) = f^{''} (x)$ .

Instead of differentiating by hand, you can use Desmos to evaluate the 2nd derivative at your critical points automatically. Simply type these two lines:

$g^{'} (- 0.947)$

$g^{'} (0.947)$

which results in:

$g^{'} (- 0.947) g^{'} (0.947) = - 5.243 = 5.243$

Analysis:

Since $g^{'} (- 0.947) < 0$ , then $x = - 0.947$ is a relative maximum on $f$
Since $g^{'} (0.947) > 0$ , then $x = 0.947$ is a relative minimum on $f$ .

Note that to use the 1st derivative test to classify extrema on $f$ given the derivative $f^{'}$ , simply observe the graph of $g$ at the critical points. For example, at $x = - 0.947$ , the graph of $g = f^{'}$ changes from positive to negative, confirming that $f$ has a relative maximum at that critical point.

Example 3: Conceptual problem

Let $f$ be a twice-differentiable function such that $f^{'} (2) = 0$ .

Let function $G$ be defined by

$G (x) = - [f (2 x)]^{2}$

If $f (2) < 0$ and $f^{''} (2) < 0$ , classify the extrema of $G (x)$ at $x = 1$ .

Note: The phrase “twice-differentiable” simply means that both $f^{'}$ and $f^{''}$ exist.

Solution

(spoiler)

To classify the extrema at $x = 1$ , we need to apply the 2nd derivative test and determine the sign of $G^{''} (1)$ . Differentiating $G$ with the chain rule,

$G^{'} (x) = - 2 f (2 x) \cdot f^{'} (2 x) \cdot 2 = - 4 f (2 x) \cdot f^{'} (2 x)$

We can confirm that $x = 1$ is a critical point since $G^{'} (1) = - 4 f (2) f^{'} (2) = 0$ (given that $f^{'} (2) = 0$ ).

Next, differentiate $G^{'}$ . using the product and chain rules:

$G^{''} (x) = - 4 (f (2 x) \cdot 2 f^{''} (2 x) + f^{'} (2 x) \cdot 2 f^{'} (2 x))$

Challenge problem

The graph of a function $y = f (x)$ has derivative

$\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$

If the line $y = - 2$ is tangent to the graph, determine the point of tangency and classify it as a relative maximum, relative minimum, or neither.

Solution

(spoiler)

The curve meets $y = - 2$ at the point of tangency $(a, - 2)$ .

Because $y = - 2$ is a horizontal tangent line, its slope is $0$ , so at the tangency point we must have

$\frac{d y}{d x} = 0.$

The first derivative

$\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$

is $0$ when the numerator equals $0$ :

$x^{2} - 4 x + 3 = 0$

$(x - 1) (x - 3) = 0$

$x = 1, 3$

So $(1, - 2)$ and $(3, - 2)$ are both points where the tangent line could be horizontal. (And the derivative is defined there because $y + 5 = - 2 + 5 = 3 \neq = 0$ .)

Now classify each point using the 2nd derivative.

The 2nd derivative is

$\frac{d ^{2} y}{d x ^{2}} = \frac{( y + 5 ) ( 2 x - 4 ) - ( x ^{2} - 4 x + 3 ) ( \frac{d y}{d x} )}{( y + 5 ) ^{2}}$

For $(1, - 2)$ :

$\frac{d ^{2} y}{d x ^{2}} = \frac{( - 2 + 5 ) ( 2 \cdot 1 - 4 ) - ( 1 ^{2} - 4 \cdot 1 + 3 ) ( 0 )}{( - 2 + 5 ) ^{2}}$

$= \frac{( 3 ) ( - 2 )}{9}$

$= - \frac{2}{3}$

Since the 2nd derivative is negative, the curve is concave down at $(1, - 2)$ , so $(1, - 2)$ is a relative max.

For $(3, - 2)$ :

$\frac{d ^{2} y}{d x ^{2}} = \frac{( - 2 + 5 ) ( 2 \cdot 3 - 4 ) - ( 3 ^{2} - 4 \cdot 3 + 3 ) ( 0 )}{( - 2 + 5 ) ^{2}}$

$= \frac{( 3 ) ( 2 )}{9}$

$= \frac{2}{3}$

Since the 2nd derivative is positive, the curve is concave up at $(3, - 2)$ , so $(3, - 2)$ is a relative min.

Essentially, the 2nd derivative test to classify extrema only works when three criteria are met:

$f (c)$ is defined.
$f^{'} (c) = 0$ .
$f^{''} (c) \neq = 0$ .

If any of those three conditions fail (namely, if $f (c)$ , $f^{'} (c)$ , or $f^{''} (c)$ is undefined, or if $f^{''} (c) = 0$ ), the test does not apply.

What you’ll learn:

How to classify relative extrema using the 2nd derivative test instead of the 1st derivative sign chart
Understanding the 2nd derivative as concavity

Once you’ve found a critical point where $f^{'} (x) = 0$ , you can often bypass a full sign chart for $f^{'} (x)$ by using the 2nd derivative test to classify extrema.

2nd derivative test for extrema

Each critical point $x = c$ can be classified by evaluating the sign of the 2nd derivative at the point, $f^{''} (c)$ :

Relative minimum: If $f^{''} (c) > 0$ , then $f (x)$ is concave up.
- Visual cue: The curve is shaped “like a cup.”
Relative maximum: If $f^{''} (c) < 0$ , then $f (x)$ is concave down.
- Visual cue: The curve is shaped “like a frown.”
The test is inconclusive if $f^{''} (x) = 0$ or is undefined.
- Use the 1st derivative test to classify the point.

Sidenote

Required condition

The 2nd derivative test only works when $f^{'} (c) = 0$ .

If $f^{'} (c)$ is undefined, then $f^{''} (c)$ is also undefined, and the test cannot be used.

As shown in the diagram below, a horizontal tangent line combined with the graph’s shape determines whether the point is a peak (relative maximum) or a valley (relative minimum).

Example 1: Classic problem

Classify the extrema of

$f (x) = \frac{1}{5} x^{5} + \frac{1}{4} x^{4}$

using the 2nd derivative test.

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = x^{4} + x^{3}$

To find the critical points, set equal to $0$ :

$x^{4} + x^{3} = 0 x^{3} (x + 1) = 0 x = 0, - 1$

Now apply the 2nd derivative test. The 2nd derivative is

$f^{''} (x) = 4 x^{3} + 3 x^{2}$

Evaluate each critical point:

$f^{''} (- 1) f^{''} (0) = - 1 = 0$

Classify each:

Since $f^{''} (- 1) < 0$ , $f (x)$ is concave down at $x = - 1$ , so it is a relative max.
Since $f^{''} (0) = 0$ , the test is inconclusive. Using the 1st derivative test,

Interval	Sign of $f^{'}$
$x < 0$	$-$
$x > 0$	$+$

Since $f^{'}$ changes from negative to positive, there is a relative min at $x = 0$ .

Example 2: Using your calculator

The derivative of a function $f$ is given by

$f^{'} (x) = 2 x^{2} - e^{c o s (x)}$

Find all critical points of $f$ and use the 2nd derivative test to classify them as relative maximums or relative minimums.

Solution

(spoiler)

Critical points of $f$ occur where $f^{'} (x) = 0$ or is undefined.

As a calculator-active question, you can simply define the given derivative of $f$ as a separate function $g$ in Desmos by typing the following:

$g (x) = 2 x^{2} - e^{c o s (x)}$

Since $g (x) = f^{'} (x)$ , note where the graph of $g$ crosses the $x$ -axis, which is where $f^{'} (x) = 0$ . These are the critical points of $x = \pm 0.947$ .

Because $g (x) = f^{'} (x)$ , then differentiating $g (x)$ gives the 2nd derivative of $f$ i.e. $g^{'} (x) = f^{''} (x)$ .

Instead of differentiating by hand, you can use Desmos to evaluate the 2nd derivative at your critical points automatically. Simply type these two lines:

$g^{'} (- 0.947)$

$g^{'} (0.947)$

which results in:

$g^{'} (- 0.947) g^{'} (0.947) = - 5.243 = 5.243$

Analysis:

Since $g^{'} (- 0.947) < 0$ , then $x = - 0.947$ is a relative maximum on $f$
Since $g^{'} (0.947) > 0$ , then $x = 0.947$ is a relative minimum on $f$ .

Example 3: Conceptual problem

Let $f$ be a twice-differentiable function such that $f^{'} (2) = 0$ .

Let function $G$ be defined by

$G (x) = - [f (2 x)]^{2}$

If $f (2) < 0$ and $f^{''} (2) < 0$ , classify the extrema of $G (x)$ at $x = 1$ .

Note: The phrase “twice-differentiable” simply means that both $f^{'}$ and $f^{''}$ exist.

Solution

(spoiler)

To classify the extrema at $x = 1$ , we need to apply the 2nd derivative test and determine the sign of $G^{''} (1)$ . Differentiating $G$ with the chain rule,

$G^{'} (x) = - 2 f (2 x) \cdot f^{'} (2 x) \cdot 2 = - 4 f (2 x) \cdot f^{'} (2 x)$

We can confirm that $x = 1$ is a critical point since $G^{'} (1) = - 4 f (2) f^{'} (2) = 0$ (given that $f^{'} (2) = 0$ ).

Next, differentiate $G^{'}$ . using the product and chain rules:

$G^{''} (x) = - 4 (f (2 x) \cdot 2 f^{''} (2 x) + f^{'} (2 x) \cdot 2 f^{'} (2 x))$

Challenge problem

The graph of a function $y = f (x)$ has derivative

$\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$

If the line $y = - 2$ is tangent to the graph, determine the point of tangency and classify it as a relative maximum, relative minimum, or neither.

Solution

(spoiler)

The curve meets $y = - 2$ at the point of tangency $(a, - 2)$ .

Because $y = - 2$ is a horizontal tangent line, its slope is $0$ , so at the tangency point we must have

$\frac{d y}{d x} = 0.$

The first derivative

$\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$

is $0$ when the numerator equals $0$ :

$x^{2} - 4 x + 3 = 0$

$(x - 1) (x - 3) = 0$

$x = 1, 3$

So $(1, - 2)$ and $(3, - 2)$ are both points where the tangent line could be horizontal. (And the derivative is defined there because $y + 5 = - 2 + 5 = 3 \neq = 0$ .)

Now classify each point using the 2nd derivative.

The 2nd derivative is

$\frac{d ^{2} y}{d x ^{2}} = \frac{( y + 5 ) ( 2 x - 4 ) - ( x ^{2} - 4 x + 3 ) ( \frac{d y}{d x} )}{( y + 5 ) ^{2}}$

For $(1, - 2)$ :

$\frac{d ^{2} y}{d x ^{2}} = \frac{( - 2 + 5 ) ( 2 \cdot 1 - 4 ) - ( 1 ^{2} - 4 \cdot 1 + 3 ) ( 0 )}{( - 2 + 5 ) ^{2}}$

$= \frac{( 3 ) ( - 2 )}{9}$

$= - \frac{2}{3}$

Since the 2nd derivative is negative, the curve is concave down at $(1, - 2)$ , so $(1, - 2)$ is a relative max.

For $(3, - 2)$ :

$\frac{d ^{2} y}{d x ^{2}} = \frac{( - 2 + 5 ) ( 2 \cdot 3 - 4 ) - ( 3 ^{2} - 4 \cdot 3 + 3 ) ( 0 )}{( - 2 + 5 ) ^{2}}$

$= \frac{( 3 ) ( 2 )}{9}$

$= \frac{2}{3}$

Since the 2nd derivative is positive, the curve is concave up at $(3, - 2)$ , so $(3, - 2)$ is a relative min.

Essentially, the 2nd derivative test to classify extrema only works when three criteria are met:

$f (c)$ is defined.
$f^{'} (c) = 0$ .
$f^{''} (c) \neq = 0$ .

If any of those three conditions fail (namely, if $f (c)$ , $f^{'} (c)$ , or $f^{''} (c)$ is undefined, or if $f^{''} (c) = 0$ ), the test does not apply.

What you’ll learn:

How to classify relative extrema using the 2nd derivative test instead of the 1st derivative sign chart
Understanding the 2nd derivative as concavity

Once you’ve found a critical point where $f^{'} (x) = 0$ , you can often bypass a full sign chart for $f^{'} (x)$ by using the 2nd derivative test to classify extrema.

2nd derivative test for extrema

Each critical point $x = c$ can be classified by evaluating the sign of the 2nd derivative at the point, $f^{''} (c)$ :

Relative minimum: If $f^{''} (c) > 0$ , then $f (x)$ is concave up.
- Visual cue: The curve is shaped “like a cup.”
Relative maximum: If $f^{''} (c) < 0$ , then $f (x)$ is concave down.
- Visual cue: The curve is shaped “like a frown.”
The test is inconclusive if $f^{''} (x) = 0$ or is undefined.
- Use the 1st derivative test to classify the point.

Sidenote

Required condition

The 2nd derivative test only works when $f^{'} (c) = 0$ .

If $f^{'} (c)$ is undefined, then $f^{''} (c)$ is also undefined, and the test cannot be used.

As shown in the diagram below, a horizontal tangent line combined with the graph’s shape determines whether the point is a peak (relative maximum) or a valley (relative minimum).

Example 1: Classic problem

Classify the extrema of

$f (x) = \frac{1}{5} x^{5} + \frac{1}{4} x^{4}$

using the 2nd derivative test.

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = x^{4} + x^{3}$

To find the critical points, set equal to $0$ :

$x^{4} + x^{3} = 0 x^{3} (x + 1) = 0 x = 0, - 1$

Now apply the 2nd derivative test. The 2nd derivative is

$f^{''} (x) = 4 x^{3} + 3 x^{2}$

Evaluate each critical point:

$f^{''} (- 1) f^{''} (0) = - 1 = 0$

Classify each:

Since $f^{''} (- 1) < 0$ , $f (x)$ is concave down at $x = - 1$ , so it is a relative max.
Since $f^{''} (0) = 0$ , the test is inconclusive. Using the 1st derivative test,

Interval	Sign of $f^{'}$
$x < 0$	$-$
$x > 0$	$+$

Since $f^{'}$ changes from negative to positive, there is a relative min at $x = 0$ .

Example 2: Using your calculator

The derivative of a function $f$ is given by

$f^{'} (x) = 2 x^{2} - e^{c o s (x)}$

Find all critical points of $f$ and use the 2nd derivative test to classify them as relative maximums or relative minimums.

Solution

(spoiler)

Critical points of $f$ occur where $f^{'} (x) = 0$ or is undefined.

As a calculator-active question, you can simply define the given derivative of $f$ as a separate function $g$ in Desmos by typing the following:

$g (x) = 2 x^{2} - e^{c o s (x)}$

Since $g (x) = f^{'} (x)$ , note where the graph of $g$ crosses the $x$ -axis, which is where $f^{'} (x) = 0$ . These are the critical points of $x = \pm 0.947$ .

Because $g (x) = f^{'} (x)$ , then differentiating $g (x)$ gives the 2nd derivative of $f$ i.e. $g^{'} (x) = f^{''} (x)$ .

Instead of differentiating by hand, you can use Desmos to evaluate the 2nd derivative at your critical points automatically. Simply type these two lines:

$g^{'} (- 0.947)$

$g^{'} (0.947)$

which results in:

$g^{'} (- 0.947) g^{'} (0.947) = - 5.243 = 5.243$

Analysis:

Since $g^{'} (- 0.947) < 0$ , then $x = - 0.947$ is a relative maximum on $f$
Since $g^{'} (0.947) > 0$ , then $x = 0.947$ is a relative minimum on $f$ .

Example 3: Conceptual problem

Let $f$ be a twice-differentiable function such that $f^{'} (2) = 0$ .

Let function $G$ be defined by

$G (x) = - [f (2 x)]^{2}$

If $f (2) < 0$ and $f^{''} (2) < 0$ , classify the extrema of $G (x)$ at $x = 1$ .

Note: The phrase “twice-differentiable” simply means that both $f^{'}$ and $f^{''}$ exist.

Solution

(spoiler)

To classify the extrema at $x = 1$ , we need to apply the 2nd derivative test and determine the sign of $G^{''} (1)$ . Differentiating $G$ with the chain rule,

$G^{'} (x) = - 2 f (2 x) \cdot f^{'} (2 x) \cdot 2 = - 4 f (2 x) \cdot f^{'} (2 x)$

We can confirm that $x = 1$ is a critical point since $G^{'} (1) = - 4 f (2) f^{'} (2) = 0$ (given that $f^{'} (2) = 0$ ).

Next, differentiate $G^{'}$ . using the product and chain rules:

$G^{''} (x) = - 4 (f (2 x) \cdot 2 f^{''} (2 x) + f^{'} (2 x) \cdot 2 f^{'} (2 x))$

Challenge problem

The graph of a function $y = f (x)$ has derivative

$\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$

If the line $y = - 2$ is tangent to the graph, determine the point of tangency and classify it as a relative maximum, relative minimum, or neither.

Solution

(spoiler)

The curve meets $y = - 2$ at the point of tangency $(a, - 2)$ .

Because $y = - 2$ is a horizontal tangent line, its slope is $0$ , so at the tangency point we must have

$\frac{d y}{d x} = 0.$

The first derivative

$\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$

is $0$ when the numerator equals $0$ :

$x^{2} - 4 x + 3 = 0$

$(x - 1) (x - 3) = 0$

$x = 1, 3$

So $(1, - 2)$ and $(3, - 2)$ are both points where the tangent line could be horizontal. (And the derivative is defined there because $y + 5 = - 2 + 5 = 3 \neq = 0$ .)

Now classify each point using the 2nd derivative.

The 2nd derivative is

$\frac{d ^{2} y}{d x ^{2}} = \frac{( y + 5 ) ( 2 x - 4 ) - ( x ^{2} - 4 x + 3 ) ( \frac{d y}{d x} )}{( y + 5 ) ^{2}}$

For $(1, - 2)$ :

$\frac{d ^{2} y}{d x ^{2}} = \frac{( - 2 + 5 ) ( 2 \cdot 1 - 4 ) - ( 1 ^{2} - 4 \cdot 1 + 3 ) ( 0 )}{( - 2 + 5 ) ^{2}}$

$= \frac{( 3 ) ( - 2 )}{9}$

$= - \frac{2}{3}$

Since the 2nd derivative is negative, the curve is concave down at $(1, - 2)$ , so $(1, - 2)$ is a relative max.

For $(3, - 2)$ :

$\frac{d ^{2} y}{d x ^{2}} = \frac{( - 2 + 5 ) ( 2 \cdot 3 - 4 ) - ( 3 ^{2} - 4 \cdot 3 + 3 ) ( 0 )}{( - 2 + 5 ) ^{2}}$

$= \frac{( 3 ) ( 2 )}{9}$

$= \frac{2}{3}$

Since the 2nd derivative is positive, the curve is concave up at $(3, - 2)$ , so $(3, - 2)$ is a relative min.

Essentially, the 2nd derivative test to classify extrema only works when three criteria are met:

$f (c)$ is defined.
$f^{'} (c) = 0$ .
$f^{''} (c) \neq = 0$ .

If any of those three conditions fail (namely, if $f (c)$ , $f^{'} (c)$ , or $f^{''} (c)$ is undefined, or if $f^{''} (c) = 0$ ), the test does not apply.