5.3.1 Concavity

Achievable AP Calculus AB

5. Analytical uses

5.3. 2nd derivative test

Our AP Calculus AB course is currently in development and is a work-in-progress.

Concavity

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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

You might recall from section 4.4 on linear approximations that concavity describes whether a graph bends upward or downward.

Once you’ve found a critical point using $f^{'} (x)$ , you can often classify it as a relative maximum or minimum using the 2nd derivative test. This lets you use $f^{''} (x)$ (concavity) instead of building a full sign chart for $f^{'} (x)$ .

2nd derivative test

Find the critical points (where $f^{'} (x) = 0$ or is undefined).
Find the 2nd derivative, $f^{''} (x)$ .
Plug each critical point into $f^{''} (x)$ :

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”), and the critical point is a relative minimum.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”), and the critical point is a relative maximum.
If $f^{''} (x) = 0$ , the 2nd derivative test is inconclusive, so you’ll need to use the 1st derivative test to classify the point.

Here’s why this works:

The 1st derivative $f^{'} (x)$ tells you the slope of the tangent line to $f (x)$ .
The 2nd derivative $f^{''} (x)$ tells you how those slopes are changing as $x$ increases.

So:

If $f^{''} (x) > 0$ , the slopes are increasing (tangent lines get “less negative” and then more positive). That corresponds to a concave-up shape.
If $f^{''} (x) < 0$ , the slopes are decreasing. That corresponds to a concave-down shape.

For example, take $f (x) = x^{2}$ .

$f^{'} (x) = 2 x$
$f^{''} (x) = 2$ (always positive, so the graph is always concave up)

At $x = - 2$ , the slope is $f^{'} (- 2) = - 4$ . At $x = - 1$ , the slope is $f^{'} (- 1) = - 2$ . The slope increased from $- 4$ to $- 2$ , which matches $f^{''} (x) > 0$ .

Now compare that to $f (x) = - x^{2}$ .

$f^{'} (x) = - 2 x$
$f^{''} (x) = - 2$ (always negative, so the graph is always concave down)

As you move left to right, the slopes decrease, matching $f^{''} (x) < 0$ .

Examples

Classify the extrema of

$f (x) = x^{3} - 3 x^{2} + 1$

Solution

The 1st derivative is

$f^{'} (x) = 3 x^{2} - 6 x$

To find the critical points, solve

$3 x^{2} - 6 x = 0$

$3 x (x - 2) = 0$

$x = 0, 2$

These are the two critical points.

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

$f^{''} (x) = 6 x - 6$

Plug in each critical point:

$f^{''} (0) = 6 (0) - 6 = - 6$

$f^{''} (2) = 6 (2) - 6 = 6$

Since $f^{''} (0) < 0$ , $f (x)$ is concave down at $x = 0$ , so $x = 0$ is a relative max.
Since $f^{''} (2) > 0$ , $f (x)$ is concave up at $x = 2$ , so $x = 2$ is a relative min.

Classify the extrema of

$f (x) = \frac{x ^{2} - 1}{x ^{3} + 1}$

Solution

(spoiler)

This function can be factored into

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

$= \frac{x - 1}{x ^{2} - x + 1}$

where $x = - 1$ is excluded from the domain.

Differentiating,

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

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

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

where $x \neq = - 1$ still.

The critical points occur when $f^{'} (x) = 0$ , which happens when the numerator is zero:

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

So $x = 0$ and $x = 2$ .

$f^{'} (x)$ is undefined when the denominator $(x^{2} - x + 1)^{2} = 0$ . There are no solutions for this (you can check the discriminant, or try solving $x^{2} - x + 1 = 0$ with the quadratic formula and see there are no real solutions).

So $x = 0$ and $x = 2$ are the only critical points.

Now use the 2nd derivative test. Differentiate $f^{'} (x)$ (quotient rule) and plug in the critical points. For calculator-active problems, you can also use technology to evaluate $f^{''} (0)$ and $f^{''} (2)$ .

For example, in Desmos:

Type $f (x) = \frac{- x ^{2} + 2 x}{( x ^{2} - x + 1 ) ^{2}}$
- This is $f^{'} (x)$ in the problem, but Desmos doesn’t support prime notation, so you enter it as a new function.
Type $f^{'} (0)$ and $f^{'} (2)$ . These outputs correspond to $f^{''} (0)$ and $f^{''} (2)$ for this problem.

You should see that:

$f^{''} (0) = 2$

$f^{''} (2) = - 0.22...$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.
Since $f^{''} (2) < 0$ , $f (x)$ is concave down at $x = 2$ , so $x = 2$ is a relative max.

Classify the extrema of

$f (x) = cos (x) - sec (x)$

in the interval $(0, 2 π)$ .

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = - sin (x) - sec (x) tan (x)$

To find the critical points, solve

$- sin (x) - sec (x) tan (x) = 0$

Rewrite in terms of sine and cosine:

$- sin (x) - \frac{1}{cos ( x )} \times \frac{sin ( x )}{cos ( x )} = 0$

$- sin (x) - \frac{sin ( x )}{cos ^{2} ( x )} = 0$

Combine into a single fraction:

$\frac{- sin ( x ) cos ^{2} ( x ) - sin ( x )}{cos ^{2} ( x )} = 0$

$= \frac{- sin ( x ) ( cos ^{2} ( x ) + 1 ))}{cos ^{2} ( x )} = 0$

A fraction is zero when its numerator is zero, so either

$cos^{2} (x) + 1 = 0$

(which has no real solutions), or

$- sin (x) = 0$

$x = 0, π, 2 π, ...$

$f^{'} (x)$ is undefined when $sec (x)$ and $tan (x)$ are undefined (equivalently, when $cos (x) = 0$ ). But those values are not in the domain of the original function, so they aren’t critical points of $f$ .

The only critical point in the interval $(0, 2 π)$ is $x = π$ .

For the 2nd derivative test,

$f^{''} (x) = - cos (x) - (sec (x) \cdot sec^{2} (x) + tan (x) \cdot sec (x) tan (x))$

$= - cos (x) - sec^{3} (x) - sec (x) tan^{2} (x)$

Classify the critical point $x = π$ :

$f^{''} (π) = 2$

Since $f^{''} (π) > 0$ , $f (x)$ is concave up at $x = π$ , so $x = π$ is a relative min.

Classify the extrema of

$f (x) = x e^{- x}$

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = x e^{- x} (- 1) + e^{- x} (1)$

$= - e^{- x} (x - 1)$

To find the critical points, solve

$- e^{- x} (x - 1) = 0$

$x = 1$

The 2nd derivative is

$f^{''} (x) = - e^{- x} (1) + (x - 1) (e^{- x})$

$= - e^{- x} (1 - (x - 1))$

$= - e^{- x} (- x + 2)$

Classify the critical point $x = 1$ :

$f^{''} (1) = - e^{- 1} (- 1 + 2) = - \frac{1}{e}$

Since $f^{''} (1) < 0$ , $f (x)$ is concave down at $x = 1$ , so $x = 1$ is a relative max.

Classify the extrema of

$f (x) = ln (x^{2} + 1)$

Solution

(spoiler)

The 1st derivative is

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

To find the critical points, solve

$\frac{2 x}{x ^{2} + 1} = 0$

$x = 0$

The 2nd derivative is

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

$= \frac{2 x ^{2} + 2 - 4 x ^{2}}{( x ^{2} + 1 ) ^{2}}$

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

Classify the critical point $x = 0$ :

$f^{''} (0) = \frac{2 - 2 ( 0 ) ^{2}}{( 0 ^{2} + 1 ) ^{2}}$

$= 2$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.

Challenge problem

A curve with derivative

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

has $y = - 2$ as a tangent line. At what point is the line tangent to the curve, and is it a relative maximum, a 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

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

From

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

the derivative is zero when the numerator is zero:

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

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

$x = 1, 3$

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.

2nd derivative test

Classifies critical points using $f^{''} (x)$ (concavity)
Steps:
- Find critical points: $f^{'} (x) = 0$ or undefined
- Compute $f^{''} (x)$ at each critical point
Interpretation:
- $f^{''} (x) > 0$ : concave up, relative minimum
- $f^{''} (x) < 0$ : concave down, relative maximum
- $f^{''} (x) = 0$ : test is inconclusive, use 1st derivative test

Concavity and derivatives

$f^{'} (x)$ : slope of tangent line
$f^{''} (x)$ : rate of change of slope (concavity)
- $f^{''} (x) > 0$ : slopes increase, graph bends upward
- $f^{''} (x) < 0$ : slopes decrease, graph bends downward

Worked examples

$f (x) = x^{3} - 3 x^{2} + 1$
- Critical points: $x = 0, 2$
- $f^{''} (0) = - 6$ (max), $f^{''} (2) = 6$ (min)
$f (x) = \frac{x ^{2} - 1}{x ^{3} + 1}$
- Critical points: $x = 0, 2$
- $f^{''} (0) = 2$ (min), $f^{''} (2) < 0$ (max)
$f (x) = cos (x) - sec (x)$ on $(0, 2 π)$
- Critical point: $x = π$
- $f^{''} (π) = 2$ (min)
$f (x) = x e^{- x}$
- Critical point: $x = 1$
- $f^{''} (1) < 0$ (max)
$f (x) = ln (x^{2} + 1)$
- Critical point: $x = 0$
- $f^{''} (0) = 2$ (min)

Challenge problem

Given $\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$ , $y = - 2$ tangent line
- Points of tangency: $(1, - 2)$ (max), $(3, - 2)$ (min)
- Classified using 2nd derivative at each point

Concavity

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

You might recall from section 4.4 on linear approximations that concavity describes whether a graph bends upward or downward.

2nd derivative test

Find the critical points (where $f^{'} (x) = 0$ or is undefined).
Find the 2nd derivative, $f^{''} (x)$ .
Plug each critical point into $f^{''} (x)$ :

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”), and the critical point is a relative minimum.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”), and the critical point is a relative maximum.
If $f^{''} (x) = 0$ , the 2nd derivative test is inconclusive, so you’ll need to use the 1st derivative test to classify the point.

Here’s why this works:

The 1st derivative $f^{'} (x)$ tells you the slope of the tangent line to $f (x)$ .
The 2nd derivative $f^{''} (x)$ tells you how those slopes are changing as $x$ increases.

So:

If $f^{''} (x) > 0$ , the slopes are increasing (tangent lines get “less negative” and then more positive). That corresponds to a concave-up shape.
If $f^{''} (x) < 0$ , the slopes are decreasing. That corresponds to a concave-down shape.

For example, take $f (x) = x^{2}$ .

$f^{'} (x) = 2 x$
$f^{''} (x) = 2$ (always positive, so the graph is always concave up)

At $x = - 2$ , the slope is $f^{'} (- 2) = - 4$ . At $x = - 1$ , the slope is $f^{'} (- 1) = - 2$ . The slope increased from $- 4$ to $- 2$ , which matches $f^{''} (x) > 0$ .

Now compare that to $f (x) = - x^{2}$ .

$f^{'} (x) = - 2 x$
$f^{''} (x) = - 2$ (always negative, so the graph is always concave down)

As you move left to right, the slopes decrease, matching $f^{''} (x) < 0$ .

Examples

Classify the extrema of

$f (x) = x^{3} - 3 x^{2} + 1$

Solution

The 1st derivative is

$f^{'} (x) = 3 x^{2} - 6 x$

To find the critical points, solve

$3 x^{2} - 6 x = 0$

$3 x (x - 2) = 0$

$x = 0, 2$

These are the two critical points.

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

$f^{''} (x) = 6 x - 6$

Plug in each critical point:

$f^{''} (0) = 6 (0) - 6 = - 6$

$f^{''} (2) = 6 (2) - 6 = 6$

Since $f^{''} (0) < 0$ , $f (x)$ is concave down at $x = 0$ , so $x = 0$ is a relative max.
Since $f^{''} (2) > 0$ , $f (x)$ is concave up at $x = 2$ , so $x = 2$ is a relative min.

Classify the extrema of

$f (x) = \frac{x ^{2} - 1}{x ^{3} + 1}$

Solution

(spoiler)

This function can be factored into

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

$= \frac{x - 1}{x ^{2} - x + 1}$

where $x = - 1$ is excluded from the domain.

Differentiating,

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

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

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

where $x \neq = - 1$ still.

The critical points occur when $f^{'} (x) = 0$ , which happens when the numerator is zero:

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

So $x = 0$ and $x = 2$ .

So $x = 0$ and $x = 2$ are the only critical points.

For example, in Desmos:

Type $f (x) = \frac{- x ^{2} + 2 x}{( x ^{2} - x + 1 ) ^{2}}$
- This is $f^{'} (x)$ in the problem, but Desmos doesn’t support prime notation, so you enter it as a new function.
Type $f^{'} (0)$ and $f^{'} (2)$ . These outputs correspond to $f^{''} (0)$ and $f^{''} (2)$ for this problem.

You should see that:

$f^{''} (0) = 2$

$f^{''} (2) = - 0.22...$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.
Since $f^{''} (2) < 0$ , $f (x)$ is concave down at $x = 2$ , so $x = 2$ is a relative max.

Classify the extrema of

$f (x) = cos (x) - sec (x)$

in the interval $(0, 2 π)$ .

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = - sin (x) - sec (x) tan (x)$

To find the critical points, solve

$- sin (x) - sec (x) tan (x) = 0$

Rewrite in terms of sine and cosine:

$- sin (x) - \frac{1}{cos ( x )} \times \frac{sin ( x )}{cos ( x )} = 0$

$- sin (x) - \frac{sin ( x )}{cos ^{2} ( x )} = 0$

Combine into a single fraction:

$\frac{- sin ( x ) cos ^{2} ( x ) - sin ( x )}{cos ^{2} ( x )} = 0$

$= \frac{- sin ( x ) ( cos ^{2} ( x ) + 1 ))}{cos ^{2} ( x )} = 0$

A fraction is zero when its numerator is zero, so either

$cos^{2} (x) + 1 = 0$

(which has no real solutions), or

$- sin (x) = 0$

$x = 0, π, 2 π, ...$

The only critical point in the interval $(0, 2 π)$ is $x = π$ .

For the 2nd derivative test,

$f^{''} (x) = - cos (x) - (sec (x) \cdot sec^{2} (x) + tan (x) \cdot sec (x) tan (x))$

$= - cos (x) - sec^{3} (x) - sec (x) tan^{2} (x)$

Classify the critical point $x = π$ :

$f^{''} (π) = 2$

Since $f^{''} (π) > 0$ , $f (x)$ is concave up at $x = π$ , so $x = π$ is a relative min.

Classify the extrema of

$f (x) = x e^{- x}$

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = x e^{- x} (- 1) + e^{- x} (1)$

$= - e^{- x} (x - 1)$

To find the critical points, solve

$- e^{- x} (x - 1) = 0$

$x = 1$

The 2nd derivative is

$f^{''} (x) = - e^{- x} (1) + (x - 1) (e^{- x})$

$= - e^{- x} (1 - (x - 1))$

$= - e^{- x} (- x + 2)$

Classify the critical point $x = 1$ :

$f^{''} (1) = - e^{- 1} (- 1 + 2) = - \frac{1}{e}$

Since $f^{''} (1) < 0$ , $f (x)$ is concave down at $x = 1$ , so $x = 1$ is a relative max.

Classify the extrema of

$f (x) = ln (x^{2} + 1)$

Solution

(spoiler)

The 1st derivative is

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

To find the critical points, solve

$\frac{2 x}{x ^{2} + 1} = 0$

$x = 0$

The 2nd derivative is

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

$= \frac{2 x ^{2} + 2 - 4 x ^{2}}{( x ^{2} + 1 ) ^{2}}$

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

Classify the critical point $x = 0$ :

$f^{''} (0) = \frac{2 - 2 ( 0 ) ^{2}}{( 0 ^{2} + 1 ) ^{2}}$

$= 2$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.

Challenge problem

A curve with derivative

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

has $y = - 2$ as a tangent line. At what point is the line tangent to the curve, and is it a relative maximum, a 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

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

From

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

the derivative is zero when the numerator is zero:

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

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

$x = 1, 3$

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.

Achievable AP Calculus AB

5. Analytical uses

5.3. 2nd derivative test

Our AP Calculus AB course is currently in development and is a work-in-progress.

Concavity

9 min read

Font

Discuss

Feedback

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

You might recall from section 4.4 on linear approximations that concavity describes whether a graph bends upward or downward.

2nd derivative test

Find the critical points (where $f^{'} (x) = 0$ or is undefined).
Find the 2nd derivative, $f^{''} (x)$ .
Plug each critical point into $f^{''} (x)$ :

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”), and the critical point is a relative minimum.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”), and the critical point is a relative maximum.
If $f^{''} (x) = 0$ , the 2nd derivative test is inconclusive, so you’ll need to use the 1st derivative test to classify the point.

Here’s why this works:

The 1st derivative $f^{'} (x)$ tells you the slope of the tangent line to $f (x)$ .
The 2nd derivative $f^{''} (x)$ tells you how those slopes are changing as $x$ increases.

So:

If $f^{''} (x) > 0$ , the slopes are increasing (tangent lines get “less negative” and then more positive). That corresponds to a concave-up shape.
If $f^{''} (x) < 0$ , the slopes are decreasing. That corresponds to a concave-down shape.

For example, take $f (x) = x^{2}$ .

$f^{'} (x) = 2 x$
$f^{''} (x) = 2$ (always positive, so the graph is always concave up)

At $x = - 2$ , the slope is $f^{'} (- 2) = - 4$ . At $x = - 1$ , the slope is $f^{'} (- 1) = - 2$ . The slope increased from $- 4$ to $- 2$ , which matches $f^{''} (x) > 0$ .

Now compare that to $f (x) = - x^{2}$ .

$f^{'} (x) = - 2 x$
$f^{''} (x) = - 2$ (always negative, so the graph is always concave down)

As you move left to right, the slopes decrease, matching $f^{''} (x) < 0$ .

Examples

Classify the extrema of

$f (x) = x^{3} - 3 x^{2} + 1$

Solution

The 1st derivative is

$f^{'} (x) = 3 x^{2} - 6 x$

To find the critical points, solve

$3 x^{2} - 6 x = 0$

$3 x (x - 2) = 0$

$x = 0, 2$

These are the two critical points.

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

$f^{''} (x) = 6 x - 6$

Plug in each critical point:

$f^{''} (0) = 6 (0) - 6 = - 6$

$f^{''} (2) = 6 (2) - 6 = 6$

Since $f^{''} (0) < 0$ , $f (x)$ is concave down at $x = 0$ , so $x = 0$ is a relative max.
Since $f^{''} (2) > 0$ , $f (x)$ is concave up at $x = 2$ , so $x = 2$ is a relative min.

Classify the extrema of

$f (x) = \frac{x ^{2} - 1}{x ^{3} + 1}$

Solution

(spoiler)

This function can be factored into

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

$= \frac{x - 1}{x ^{2} - x + 1}$

where $x = - 1$ is excluded from the domain.

Differentiating,

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

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

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

where $x \neq = - 1$ still.

The critical points occur when $f^{'} (x) = 0$ , which happens when the numerator is zero:

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

So $x = 0$ and $x = 2$ .

So $x = 0$ and $x = 2$ are the only critical points.

For example, in Desmos:

Type $f (x) = \frac{- x ^{2} + 2 x}{( x ^{2} - x + 1 ) ^{2}}$
- This is $f^{'} (x)$ in the problem, but Desmos doesn’t support prime notation, so you enter it as a new function.
Type $f^{'} (0)$ and $f^{'} (2)$ . These outputs correspond to $f^{''} (0)$ and $f^{''} (2)$ for this problem.

You should see that:

$f^{''} (0) = 2$

$f^{''} (2) = - 0.22...$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.
Since $f^{''} (2) < 0$ , $f (x)$ is concave down at $x = 2$ , so $x = 2$ is a relative max.

Classify the extrema of

$f (x) = cos (x) - sec (x)$

in the interval $(0, 2 π)$ .

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = - sin (x) - sec (x) tan (x)$

To find the critical points, solve

$- sin (x) - sec (x) tan (x) = 0$

Rewrite in terms of sine and cosine:

$- sin (x) - \frac{1}{cos ( x )} \times \frac{sin ( x )}{cos ( x )} = 0$

$- sin (x) - \frac{sin ( x )}{cos ^{2} ( x )} = 0$

Combine into a single fraction:

$\frac{- sin ( x ) cos ^{2} ( x ) - sin ( x )}{cos ^{2} ( x )} = 0$

$= \frac{- sin ( x ) ( cos ^{2} ( x ) + 1 ))}{cos ^{2} ( x )} = 0$

A fraction is zero when its numerator is zero, so either

$cos^{2} (x) + 1 = 0$

(which has no real solutions), or

$- sin (x) = 0$

$x = 0, π, 2 π, ...$

The only critical point in the interval $(0, 2 π)$ is $x = π$ .

For the 2nd derivative test,

$f^{''} (x) = - cos (x) - (sec (x) \cdot sec^{2} (x) + tan (x) \cdot sec (x) tan (x))$

$= - cos (x) - sec^{3} (x) - sec (x) tan^{2} (x)$

Classify the critical point $x = π$ :

$f^{''} (π) = 2$

Since $f^{''} (π) > 0$ , $f (x)$ is concave up at $x = π$ , so $x = π$ is a relative min.

Classify the extrema of

$f (x) = x e^{- x}$

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = x e^{- x} (- 1) + e^{- x} (1)$

$= - e^{- x} (x - 1)$

To find the critical points, solve

$- e^{- x} (x - 1) = 0$

$x = 1$

The 2nd derivative is

$f^{''} (x) = - e^{- x} (1) + (x - 1) (e^{- x})$

$= - e^{- x} (1 - (x - 1))$

$= - e^{- x} (- x + 2)$

Classify the critical point $x = 1$ :

$f^{''} (1) = - e^{- 1} (- 1 + 2) = - \frac{1}{e}$

Since $f^{''} (1) < 0$ , $f (x)$ is concave down at $x = 1$ , so $x = 1$ is a relative max.

Classify the extrema of

$f (x) = ln (x^{2} + 1)$

Solution

(spoiler)

The 1st derivative is

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

To find the critical points, solve

$\frac{2 x}{x ^{2} + 1} = 0$

$x = 0$

The 2nd derivative is

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

$= \frac{2 x ^{2} + 2 - 4 x ^{2}}{( x ^{2} + 1 ) ^{2}}$

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

Classify the critical point $x = 0$ :

$f^{''} (0) = \frac{2 - 2 ( 0 ) ^{2}}{( 0 ^{2} + 1 ) ^{2}}$

$= 2$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.

Challenge problem

A curve with derivative

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

has $y = - 2$ as a tangent line. At what point is the line tangent to the curve, and is it a relative maximum, a 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

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

From

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

the derivative is zero when the numerator is zero:

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

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

$x = 1, 3$

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.

2nd derivative test

Classifies critical points using $f^{''} (x)$ (concavity)
Steps:
- Find critical points: $f^{'} (x) = 0$ or undefined
- Compute $f^{''} (x)$ at each critical point
Interpretation:
- $f^{''} (x) > 0$ : concave up, relative minimum
- $f^{''} (x) < 0$ : concave down, relative maximum
- $f^{''} (x) = 0$ : test is inconclusive, use 1st derivative test

Concavity and derivatives

$f^{'} (x)$ : slope of tangent line
$f^{''} (x)$ : rate of change of slope (concavity)
- $f^{''} (x) > 0$ : slopes increase, graph bends upward
- $f^{''} (x) < 0$ : slopes decrease, graph bends downward

Worked examples

$f (x) = x^{3} - 3 x^{2} + 1$
- Critical points: $x = 0, 2$
- $f^{''} (0) = - 6$ (max), $f^{''} (2) = 6$ (min)
$f (x) = \frac{x ^{2} - 1}{x ^{3} + 1}$
- Critical points: $x = 0, 2$
- $f^{''} (0) = 2$ (min), $f^{''} (2) < 0$ (max)
$f (x) = cos (x) - sec (x)$ on $(0, 2 π)$
- Critical point: $x = π$
- $f^{''} (π) = 2$ (min)
$f (x) = x e^{- x}$
- Critical point: $x = 1$
- $f^{''} (1) < 0$ (max)
$f (x) = ln (x^{2} + 1)$
- Critical point: $x = 0$
- $f^{''} (0) = 2$ (min)

Challenge problem

Given $\frac{d y}{d x} = \frac{x ^{2} - 4 x + 3}{y + 5}$ , $y = - 2$ tangent line
- Points of tangency: $(1, - 2)$ (max), $(3, - 2)$ (min)
- Classified using 2nd derivative at each point

Concavity

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

You might recall from section 4.4 on linear approximations that concavity describes whether a graph bends upward or downward.

2nd derivative test

Find the critical points (where $f^{'} (x) = 0$ or is undefined).
Find the 2nd derivative, $f^{''} (x)$ .
Plug each critical point into $f^{''} (x)$ :

If $f^{''} (x) > 0$ , then $f (x)$ is concave up (shaped “like a cup”), and the critical point is a relative minimum.
If $f^{''} (x) < 0$ , then $f (x)$ is concave down (shaped “like a frown”), and the critical point is a relative maximum.
If $f^{''} (x) = 0$ , the 2nd derivative test is inconclusive, so you’ll need to use the 1st derivative test to classify the point.

Here’s why this works:

The 1st derivative $f^{'} (x)$ tells you the slope of the tangent line to $f (x)$ .
The 2nd derivative $f^{''} (x)$ tells you how those slopes are changing as $x$ increases.

So:

If $f^{''} (x) > 0$ , the slopes are increasing (tangent lines get “less negative” and then more positive). That corresponds to a concave-up shape.
If $f^{''} (x) < 0$ , the slopes are decreasing. That corresponds to a concave-down shape.

For example, take $f (x) = x^{2}$ .

$f^{'} (x) = 2 x$
$f^{''} (x) = 2$ (always positive, so the graph is always concave up)

At $x = - 2$ , the slope is $f^{'} (- 2) = - 4$ . At $x = - 1$ , the slope is $f^{'} (- 1) = - 2$ . The slope increased from $- 4$ to $- 2$ , which matches $f^{''} (x) > 0$ .

Now compare that to $f (x) = - x^{2}$ .

$f^{'} (x) = - 2 x$
$f^{''} (x) = - 2$ (always negative, so the graph is always concave down)

As you move left to right, the slopes decrease, matching $f^{''} (x) < 0$ .

Examples

Classify the extrema of

$f (x) = x^{3} - 3 x^{2} + 1$

Solution

The 1st derivative is

$f^{'} (x) = 3 x^{2} - 6 x$

To find the critical points, solve

$3 x^{2} - 6 x = 0$

$3 x (x - 2) = 0$

$x = 0, 2$

These are the two critical points.

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

$f^{''} (x) = 6 x - 6$

Plug in each critical point:

$f^{''} (0) = 6 (0) - 6 = - 6$

$f^{''} (2) = 6 (2) - 6 = 6$

Since $f^{''} (0) < 0$ , $f (x)$ is concave down at $x = 0$ , so $x = 0$ is a relative max.
Since $f^{''} (2) > 0$ , $f (x)$ is concave up at $x = 2$ , so $x = 2$ is a relative min.

Classify the extrema of

$f (x) = \frac{x ^{2} - 1}{x ^{3} + 1}$

Solution

(spoiler)

This function can be factored into

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

$= \frac{x - 1}{x ^{2} - x + 1}$

where $x = - 1$ is excluded from the domain.

Differentiating,

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

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

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

where $x \neq = - 1$ still.

The critical points occur when $f^{'} (x) = 0$ , which happens when the numerator is zero:

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

So $x = 0$ and $x = 2$ .

So $x = 0$ and $x = 2$ are the only critical points.

For example, in Desmos:

Type $f (x) = \frac{- x ^{2} + 2 x}{( x ^{2} - x + 1 ) ^{2}}$
- This is $f^{'} (x)$ in the problem, but Desmos doesn’t support prime notation, so you enter it as a new function.
Type $f^{'} (0)$ and $f^{'} (2)$ . These outputs correspond to $f^{''} (0)$ and $f^{''} (2)$ for this problem.

You should see that:

$f^{''} (0) = 2$

$f^{''} (2) = - 0.22...$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.
Since $f^{''} (2) < 0$ , $f (x)$ is concave down at $x = 2$ , so $x = 2$ is a relative max.

Classify the extrema of

$f (x) = cos (x) - sec (x)$

in the interval $(0, 2 π)$ .

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = - sin (x) - sec (x) tan (x)$

To find the critical points, solve

$- sin (x) - sec (x) tan (x) = 0$

Rewrite in terms of sine and cosine:

$- sin (x) - \frac{1}{cos ( x )} \times \frac{sin ( x )}{cos ( x )} = 0$

$- sin (x) - \frac{sin ( x )}{cos ^{2} ( x )} = 0$

Combine into a single fraction:

$\frac{- sin ( x ) cos ^{2} ( x ) - sin ( x )}{cos ^{2} ( x )} = 0$

$= \frac{- sin ( x ) ( cos ^{2} ( x ) + 1 ))}{cos ^{2} ( x )} = 0$

A fraction is zero when its numerator is zero, so either

$cos^{2} (x) + 1 = 0$

(which has no real solutions), or

$- sin (x) = 0$

$x = 0, π, 2 π, ...$

The only critical point in the interval $(0, 2 π)$ is $x = π$ .

For the 2nd derivative test,

$f^{''} (x) = - cos (x) - (sec (x) \cdot sec^{2} (x) + tan (x) \cdot sec (x) tan (x))$

$= - cos (x) - sec^{3} (x) - sec (x) tan^{2} (x)$

Classify the critical point $x = π$ :

$f^{''} (π) = 2$

Since $f^{''} (π) > 0$ , $f (x)$ is concave up at $x = π$ , so $x = π$ is a relative min.

Classify the extrema of

$f (x) = x e^{- x}$

Solution

(spoiler)

The 1st derivative is

$f^{'} (x) = x e^{- x} (- 1) + e^{- x} (1)$

$= - e^{- x} (x - 1)$

To find the critical points, solve

$- e^{- x} (x - 1) = 0$

$x = 1$

The 2nd derivative is

$f^{''} (x) = - e^{- x} (1) + (x - 1) (e^{- x})$

$= - e^{- x} (1 - (x - 1))$

$= - e^{- x} (- x + 2)$

Classify the critical point $x = 1$ :

$f^{''} (1) = - e^{- 1} (- 1 + 2) = - \frac{1}{e}$

Since $f^{''} (1) < 0$ , $f (x)$ is concave down at $x = 1$ , so $x = 1$ is a relative max.

Classify the extrema of

$f (x) = ln (x^{2} + 1)$

Solution

(spoiler)

The 1st derivative is

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

To find the critical points, solve

$\frac{2 x}{x ^{2} + 1} = 0$

$x = 0$

The 2nd derivative is

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

$= \frac{2 x ^{2} + 2 - 4 x ^{2}}{( x ^{2} + 1 ) ^{2}}$

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

Classify the critical point $x = 0$ :

$f^{''} (0) = \frac{2 - 2 ( 0 ) ^{2}}{( 0 ^{2} + 1 ) ^{2}}$

$= 2$

Since $f^{''} (0) > 0$ , $f (x)$ is concave up at $x = 0$ , so $x = 0$ is a relative min.

Challenge problem

A curve with derivative

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

has $y = - 2$ as a tangent line. At what point is the line tangent to the curve, and is it a relative maximum, a 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

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

From

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

the derivative is zero when the numerator is zero:

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

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

$x = 1, 3$

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.