Derivative Test

113. Derivative Test

Find critical points by solving $𝑓^{'} (𝑥) = 0$

$1^{st}$ Derivative	$+ ⟶ -$	Local Max	Concave Down
$1^{st}$ Derivative	$- ⟶ +$	Local Min	Concave Up
$2^{nd}$ Derivative	$𝑓^{″} (𝑐) < 0$	Local Max	Concave Down
$2^{nd}$ Derivative	$𝑓^{″} (𝑐) > 0$	Local Min	Concave Up

113.0.1. First Derivative Test

If $𝑓^{'} (𝑥)$ changes sign around a critical point $𝑐$ , we can determine if $𝑓 (𝑐)$ is a local maximum or minimum:

If $𝑓^{'} (𝑥)$ changes from positive to negative at $𝑐$ , then $𝑓 (𝑐)$ is a local maximum
If $𝑓^{'} (𝑥)$ changes from negative to positive at $𝑐$ , then $𝑓 (𝑐)$ is a local minimum
If $𝑓^{'} (𝑥)$ does not change sign, $𝑓 (𝑐)$ is not a local extremum

Example

If $𝑓^{'} (𝑥)$ changes from negative to positive at $𝑐$ , then $𝑓 (𝑐)$ is a local minimum

Suppose:

𝑓 (𝑥) = 𝑥^{2}, 𝑓^{'} (𝑥) = 2 𝑥

Step 1. Find critical point

Solve $𝑓^{'} (𝑥) = 0$ :

2 𝑥 = 0 ⟹ 𝑥 = 0

So, $𝑥 = 0$ is the only critical point

Step 2. Test sign of $𝑓^{'} (𝑥)$ around $𝑥 = 0$

Left: $𝑥 = - 1$

𝑓^{'} (- 1) = 2 (- 1) = - 2 < 0 ⟹ 𝑓 is 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Right: $𝑥 = 1$ :

𝑓^{'} (1) = 2 (1) = 2 > 0 ⟹ 𝑓 is 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Step 3. Interpretation

Since $𝑓^{'} (𝑥)$ changes from negativetopositive at $𝑥 = 0$ ,

𝑓 (0) = 0 ⟹ 𝐥𝐨𝐜𝐚𝐥 𝐦𝐢𝐧𝐢𝐦𝐮𝐦

Example

If $𝑓^{'} (𝑥)$ changes from positive to negative at $𝑐$ , then $𝑓 (𝑐)$ is a local maximum

Suppose:

𝑓 (𝑥) = - 𝑥^{2}, 𝑓^{'} (𝑥) = - 2 𝑥

Step 1. Find critical point

Solve $𝑓^{'} (𝑥) = 0$

- 2 𝑥 = 0 ⟹ 𝑥 = 0

So, $𝑥 = 0$ is the only critical point

Step 2. Test sign of $𝑓^{'} (𝑥)$ around $𝑥 = 0$

Left: $𝑥 = - 1$

𝑓^{'} (- 1) = - 2 (- 1) = 2 > 0 ⟹ 𝑓 is 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Right: $𝑥 = 1$

𝑓^{'} (1) = - 2 (1) = - 2 < 0 ⟹ 𝑓 is 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Step 3. Interpretation

Since $𝑓^{'} (𝑥)$ changes from positivetonegative at $𝑥 = 0$ ,

𝑓 (0) = 0 ⟹ 𝐥𝐨𝐜𝐚𝐥 𝐦𝐚𝐱𝐢𝐦𝐮𝐦

Example

If $𝑓^{'} (𝑥)$ does not change sign, $𝑓 (𝑐)$ is not a local extremum

Suppose

𝑓 (𝑥) = 𝑥^{3}, 𝑓^{'} (𝑥) = 3 𝑥^{2}

Step 1. Find critical point

Solve $𝑓^{'} (𝑥) = 0$

3 𝑥^{2} = 0 ⟹ 𝑥 = 0

So, $𝑥 = 0$ is the only critical point

Step 2. Test sign of $𝑓^{'} (𝑥)$ around $𝑥 = 0$

Left: $𝑥 = - 1$

𝑓^{'} (- 1) = 3 {(- 1)}^{2} = 3 > 0 ⟹ 𝑓 is 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Right: $𝑥 = 1$

𝑓^{'} (1) = 3 {(1)}^{2} = 3 > 0 ⟹ 𝑓 is 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Step 3. Interpretation

Since $𝑓^{'} (𝑥)$ does not change sign at $𝑥 = 0$ ,

𝑓 (0) = 0 ⟹ 𝐧𝐨𝐭 𝐚 𝐥𝐨𝐜𝐚𝐥 𝐞𝐱𝐭𝐫𝐞𝐦𝐮𝐦

but rather a point of inflection

113.0.2. Second Derivative Test

If $𝑓^{″} (𝑥)$ is continuous near a critical point $𝑐$ , and $𝑓^{'} (𝑐) = 0$ , then:

If $𝑓^{″} (𝑐) > 0$ , then $𝑓 (𝑐)$ is a local minimum (concave up)
If $𝑓^{″} (𝑐) < 0$ , then $𝑓 (𝑐)$ is a local maximum (concave down)
If $𝑓^{″} (𝑐) = 0$ , the test is inconclusive — use the first derivative test or other methods

Example

If $𝑓^{″} (𝑐) > 0$ , then $𝑓 (𝑐)$ is a local minimum (concave up)

Suppose

𝑓 (𝑥) = 𝑥^{2}, 𝑓^{'} (𝑥) = 2 𝑥, 𝑓^{″} (𝑥) = 2

Step 1. Find critical points

Solve $𝑓^{'} (𝑥) = 0$

2 𝑥 = 0 ⟹ 𝑥 = 0

So, $𝑥 = 0$ is the only critical point

Step 2. Evaluate the second derivative at $𝑥 = 0$

𝑓^{″} (0) = 2 > 0

Since the second derivative is positive, the function is concave up near $𝑥 = 0$

Step 3. Interpretation

𝑓 (0) = 0 ⟹ 𝐥𝐨𝐜𝐚𝐥 𝐦𝐢𝐧𝐢𝐦𝐮𝐦

Example

If $𝑓^{″} (𝑐) < 0$ , then $𝑓 (𝑐)$ is a local maximum (concave down)

Suppose

𝑓 (𝑥) = - 𝑥^{2}, 𝑓^{'} (𝑥) = - 2 𝑥, 𝑓^{″} (𝑥) = - 2

Step 1. Find critical points

Solve $𝑓^{'} (𝑥) = 0$

- 2 𝑥 = 0 ⟹ 𝑥 = 0

So, $𝑥 = 0$ is the only critical point

Step 2. Evaluate the second derivative at $𝑥 = 0$

𝑓^{″} (0) = - 2 < 0

Since the second derivative is negative, the function is concave down near $𝑥 = 0$

Step 3. Interpretation

𝑓 (0) = 0 ⟹ 𝐥𝐨𝐜𝐚𝐥 𝐦𝐚𝐱𝐢𝐦𝐮𝐦

Example

If $𝑓^{″} (𝑐) = 0$ , the test is inconclusive — use the first derivative test or other methods

Suppose

𝑓 (𝑥) = 𝑥^{3}, 𝑓^{'} (𝑥) = 3 𝑥^{2}, 𝑓^{″} (𝑥) = 6 𝑥

Step 1. Find critical points

Solve $𝑓^{'} (𝑥) = 0$

3 𝑥^{2} = 0 ⟹ 𝑥 = 0

So, $𝑥 = 0$ is the only critical point

Step 2. Evaluate the second derivative at $𝑥 = 0$

𝑓^{″} (0) = 6 (0) = 0

Since the second derivative is 0, the test in inconclusive

Step 3. Use another method

Test sign of $𝑓^{'} (𝑥)$ around $𝑥 = 0$

Left: $𝑥 = - 1$

𝑓^{'} (- 1) = 3 {(- 1)}^{2} = 3 > 0 ⟹ 𝑓 is 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Right: $𝑥 = 1$

𝑓^{'} (1) = 3 {(1)}^{2} = 3 > 0 ⟹ 𝑓 is 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠

Step 4. Interpretation

Since $𝑓^{'} (𝑥)$ does not change sign at $𝑥 = 0$ ,

𝑓 (0) = 0 ⟹ 𝐧𝐨𝐭 𝐚 𝐥𝐨𝐜𝐚𝐥 𝐞𝐱𝐭𝐫𝐞𝐦𝐮𝐦

but rather a point of inflection