5.7 - Using the Second Derivative Test to Determine Extrema

Understanding the Test

The second derivative test uses the value of

f''(x)

at a critical point to classify it as a max or min.

5.7 - Using the Second Derivative Test to Determine Extrema

At a maximum:

f'(x) = 0

(horizontal tangent) and

f''(x) < 0

(concave down)

At a minimum:

f'(x) = 0

(horizontal tangent) and

f''(x) > 0

(concave up)

The Test

Suppose

f'(c) = 0

and

f''(c)

exists.

\text{If } f''(c) < 0, \text{ then } f \text{ has a relative maximum at } x = c

\text{If } f''(c) > 0, \text{ then } f \text{ has a relative minimum at } x = c

Important: the second derivative test only works when

f'(c) = 0

. It does not apply at critical points where the derivative doesn’t exist.

Why it works: concave down with a horizontal tangent means the function curves downward away from that point, so it’s a max. Concave up with a horizontal tangent means it curves upward, so it’s a min.

The Procedure

Find $f'(x)$ and solve $f'(x) = 0$ to get critical points

Find $f''(x)$

Evaluate $f''(x)$ at each critical point where $f'(c) = 0$

Classify:

$f''(c) < 0$ means relative maximum

$f''(c) > 0$ means relative minimum

$f''(c) = 0$ means the test is inconclusive, use the first derivative test instead

Example 1

Find and classify all relative extrema of

f(x) = x^4 - 2x^2 + 3

Step 1: Critical points.

f'(x) = 4x^3 - 4x = 4x(x-1)(x+1)

x = -1, 0, 1

Step 2: Second derivative.

f''(x) = 12x^2 - 4

Step 3: Evaluate.

f''(-1) = 12 - 4 = 8 > 0

f''(0) = -4 < 0

f''(1) = 12 - 4 = 8 > 0

Step 4: Classify.

Relative maximum at

x = 0

because

f'(0) = 0

and

f''(0) = -4 < 0

Relative minimum at

x = -1

because

f'(-1) = 0

and

f''(-1) = 8 > 0

Relative minimum at

x = 1

because

f'(1) = 0

and

f''(1) = 8 > 0

Justification format: always state both conditions, that the first derivative is zero AND that the second derivative is positive or negative.

Example 2

Find and classify all relative extrema of

g(x) = \sqrt{2}\,x - 2\cos(x)

[0, 2\pi]

Step 1: Critical points.

g'(x) = \sqrt{2} + 2\sin(x) = 0 \quad \Rightarrow \quad \sin(x) = -\frac{\sqrt{2}}{2}

x = \frac{5\pi}{4}, \frac{7\pi}{4}

Step 2: Second derivative.

g''(x) = 2\cos(x)

Step 3: Evaluate.

g''\left(\frac{5\pi}{4}\right) = 2\left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} < 0

g''\left(\frac{7\pi}{4}\right) = 2\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2} > 0

Step 4: Classify.

Relative maximum at

x = \frac{5\pi}{4}

because

g'\left(\frac{5\pi}{4}\right) = 0

and

g''\left(\frac{5\pi}{4}\right) < 0

Relative minimum at

x = \frac{7\pi}{4}

because

g'\left(\frac{7\pi}{4}\right) = 0

and

g''\left(\frac{7\pi}{4}\right) > 0

Example 3: Product with an Exponential

Find and classify all extrema of

f(x) = \left(1 - \dfrac{x}{4}\right)e^{x/4}

Step 1: First derivative (product rule).

f'(x) = \left(-\frac{1}{4}\right)e^{x/4} + \left(1 - \frac{x}{4}\right)e^{x/4}\cdot\frac{1}{4}

Factor out

e^{x/4}

f'(x) = e^{x/4}\left[-\frac{1}{4} + \frac{1}{4} - \frac{x}{16}\right] = e^{x/4}\left(-\frac{x}{16}\right)

Set it to zero. Since

e^{x/4}

is never zero:

-\frac{x}{16} = 0 \quad \Rightarrow \quad x = 0

Step 2: Second derivative (product rule again).

f''(x) = \left(-\frac{1}{16}\right)e^{x/4} + \left(-\frac{x}{16}\right)e^{x/4}\cdot\frac{1}{4} = e^{x/4}\left(-\frac{1}{16} - \frac{x}{64}\right)

Step 3: Evaluate at $x = 0$ .

f''(0) = e^{0}\left(-\frac{1}{16}\right) = -\frac{1}{16} < 0

Step 4: Classify.

Relative maximum at

x = 0

because

f'(0) = 0

and

f''(0) < 0

And since

x = 0

is the only critical point, this is actually an absolute maximum.

Absolute Extrema with One Critical Point

Special case: if a function is continuous on an interval and has only one critical point there, and that point is a relative extremum, then it has to be an absolute extremum on that interval, not just a relative one.

The reasoning: if there’s only one critical point and it’s a maximum, there are no other peaks anywhere, so it’s the highest point on the whole function. Same logic for a minimum.

Continuity matters here. For example,

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

has only one critical point, a relative minimum at

(1, 3)

. But

f

isn’t continuous at

x = 0

, and just to the left of

0

it drops toward

-\infty

, so

(1, 3)

is not the absolute minimum.

So if

f(x)

has only one critical point at

x = c

If $f'(c) = 0$ and $f''(c) < 0$ , then $x = c$ is an absolute maximum

If $f'(c) = 0$ and $f''(c) > 0$ , then $x = c$ is an absolute minimum

When the Second Derivative Test Fails

f''(c) = 0

, the test is inconclusive and you have to fall back on the first derivative test.

Example: for

f(x) = x^4

$f'(0) = 0$ , so there’s a critical point at $x = 0$

$f''(0) = 0$ , so the test fails

Use the first derivative test instead: $f'(x)$ changes from negative to positive, so there’s a minimum at $x = 0$

Key Takeaways

Second derivative test: if $f'(c) = 0$ and $f''(c) < 0$ it’s a relative max; if $f''(c) > 0$ it’s a relative min

The first derivative must equal zero, not just fail to exist

Concave down at a horizontal tangent means maximum

Concave up at a horizontal tangent means minimum

Justification needs both parts: $f'(c) = 0$ AND the sign of $f''(c)$

Continuous on an interval with only one critical point there, and it’s an extremum? Then it’s an absolute extremum on that interval

The test fails when $f''(c) = 0$ , so use the first derivative test

Often easier than the first derivative test, since you only evaluate $f''$ at the critical points instead of testing whole intervals

Back Next