Concavity | 2nd derivative test | Analytical uses

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 in section 4.4 on linear approximations that the term concavity (whether the graph bends upward or downward) was briefly mentioned.

After finding a critical point with $f^{'} (x)$ , another way to classify it as a relative maximum or minimum is by using the 2nd derivative test with $f^{''} (x)$ to determine the concavity at that point instead of creating a sign diagram or chart.

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:

Since the 2nd derivative is found by differentiating $f^{'} (x)$ , it tells you how the $f^{'} (x)$ changes. In turn, this tells you how the slopes of the tangent lines to the original function change.

Think of $f (x) = x^{2}$ . $f^{'} (x) = 2 x$ and $f^{''} (x) = 2$ (always concave up).

If you draw a tangent line to $f (x)$ at $x = - 2$ , it will have a slope of $- 4$ . Draw a tangent line at $x = - 1$ and it will have a slope of $- 2$ . The slope increases. If $f^{'} (x)$ has increased over time, it means that $f^{''} (x) > 0$ , in the same way that $f (x)$ increasing means that $f^{'} (x) > 0$ .

Repeat this for $f (x) = - x^{2}$ , which is a function that is always concave down ( $f^{''} (x) < 0$ ) to see how the slopes decrease over time (becoming less positive and then more negative as you move from left to right).

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,

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

$3 x (x - 2) = 0$

$x = 0, 2$

These are the two critical points

Now we’ll use the 2nd derivative test. The 2nd derivative is

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

Plugging in each of the critical points,

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

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

Since $f^{''} (0) < 0$ , the function $f (x)$ is concave down there, making $x = 0$ a relative max.

$f^{''} (2) > 0$ which means the function $f (x)$ is concave up there. 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 $f^{'} (x) = 0$ , or when the numerator $- x^{2} + 2 x = 0.$ Solving, $x = 0$ and $2$ .

$f^{'} (x)$ is undefined when the denominator $(x^{2} - x + 1)^{2} = 0$ . There are no solutions for this (use the discriminant or try to solve $x^{2} - x + 1 = 0$ with the quadratic equation to find no real solutions).

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

Now we’ll use the 2nd derivative test. Differentiate $f^{'} (x)$ with the quotient rule and plug in the critical points, or for calculator-active problems, you can enter in $f^{'} (x)$ as the function to be differentiated. This is what you would do 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 the prime notation so it has to be done this way.
Type $f^{'} (0)$ and $f^{'} (2)$ . These will give you $f^{''} (0)$ and $f^{''} (2)$ for our specific problem.

You should see that:

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

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

Since $f^{''} (0) > 0$ , the function $f (x)$ is concave up there, making $x = 0$ a relative min.

$f^{''} (2) < 0$ which means the function $f (x)$ is concave down there. 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,

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

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

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

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

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

Either

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

which is an equation that has no solutions, or

$- sin (x) = 0$

$x = 0, π, 2 π, ...$

$f^{'} (x)$ is undefined when $sec (x)$ and $tan (x)$ are undefined (or when $cos (x) = 0$ ), but since those values were excluded in the original function, they aren’t critical points.

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

For the 2nd derivative test,

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

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

Classifying the critical point $x = π$ ,

$f^{''} (π) = 2$

$f^{''} (π) > 0$ which means the function $f (x)$ is concave up at that point, and $x = 1$ 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,

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

$x = 1$

The 2nd derivative is

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

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

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

Classifying the critical point $x = 1$ ,

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

$f^{''} (1) < 0$ which means the function $f (x)$ is concave down at that point, 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,

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

$x = 0$

The 2nd derivative is

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

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

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

Classifying 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 that point, making $x = 2$ 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 with a slope of $0$ , $\frac{d y}{d x} = 0$ . Solving $x^{2} - 4 x + 3 = 0$ ,

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

$x = 1, 3$

Both $(1, - 2)$ and $(3, - 2)$ are critical points where the 1st derivative is $0$ (we know the function is defined at those points because it meets the horizontal tangent line).

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 the point $(1, - 2) \to$ 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 the point $(3, - 2) \to$ relative min.

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 in section 4.4 on linear approximations that the term concavity (whether the graph bends upward or downward) was briefly mentioned.

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:

Think of $f (x) = x^{2}$ . $f^{'} (x) = 2 x$ and $f^{''} (x) = 2$ (always concave up).

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,

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

$3 x (x - 2) = 0$

$x = 0, 2$

These are the two critical points

Now we’ll use the 2nd derivative test. The 2nd derivative is

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

Plugging in each of the critical points,

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

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

Since $f^{''} (0) < 0$ , the function $f (x)$ is concave down there, making $x = 0$ a relative max.

$f^{''} (2) > 0$ which means the function $f (x)$ is concave up there. 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 $f^{'} (x) = 0$ , or when the numerator $- x^{2} + 2 x = 0.$ Solving, $x = 0$ and $2$ .

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

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 the prime notation so it has to be done this way.
Type $f^{'} (0)$ and $f^{'} (2)$ . These will give you $f^{''} (0)$ and $f^{''} (2)$ for our specific problem.

You should see that:

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

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

Since $f^{''} (0) > 0$ , the function $f (x)$ is concave up there, making $x = 0$ a relative min.

$f^{''} (2) < 0$ which means the function $f (x)$ is concave down there. 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,

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

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

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

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

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

Either

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

which is an equation that has no solutions, or

$- sin (x) = 0$

$x = 0, π, 2 π, ...$

$f^{'} (x)$ is undefined when $sec (x)$ and $tan (x)$ are undefined (or when $cos (x) = 0$ ), but since those values were excluded in the original function, they aren’t critical points.

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

For the 2nd derivative test,

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

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

Classifying the critical point $x = π$ ,

$f^{''} (π) = 2$

$f^{''} (π) > 0$ which means the function $f (x)$ is concave up at that point, and $x = 1$ 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,

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

$x = 1$

The 2nd derivative is

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

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

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

Classifying the critical point $x = 1$ ,

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

$f^{''} (1) < 0$ which means the function $f (x)$ is concave down at that point, 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,

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

$x = 0$

The 2nd derivative is

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

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

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

Classifying 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 that point, making $x = 2$ 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 with a slope of $0$ , $\frac{d y}{d x} = 0$ . Solving $x^{2} - 4 x + 3 = 0$ ,

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

$x = 1, 3$

Both $(1, - 2)$ and $(3, - 2)$ are critical points where the 1st derivative is $0$ (we know the function is defined at those points because it meets the horizontal tangent line).

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 the point $(1, - 2) \to$ 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 the point $(3, - 2) \to$ relative min.