5.3.2 Inflection points

Achievable AP Calculus AB

5. Analytical uses

5.3. 2nd derivative test

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Inflection points

5 min read

Font

Discuss

Feedback

What you’ll learn:

Points of inflection: Identify where a function changes concavity.
Intervals of concavity: Determine where a graph bends upward or downward based on the sign of the 2nd derivative.

An inflection point (or point of inflection) occurs when the graph of a function changes concavity, from concave up to concave down or vice versa.

A potential inflection point occurs where the second derivative $f^{''} (x) = 0$ or is undefined. To confirm it is an inflection point, the sign of $f^{''} (x)$ must change across that point.

Finding points of inflection:

Find all values of $x$ where $f^{''} (x) = 0$ or is undefined. These values must lie within the domain of the original function.
Create a sign chart for $f^{''} (x)$ using these values as interval boundaries.
If $f^{''} (x)$ changes sign across a boundary, that boundary point is an inflection point.

Example 1: Analytical

Find the points of inflection and determine the intervals of concavity for

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

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$e^{- x^{2}} (4 x^{2} - 2) = 0$

Because $e^{- x^{2}} > 0$ for all $x$ , solve:

$4 x^{2} - 2 x^{2} x = 0 = \frac{1}{2} = \pm \frac{2}{2}$

4. Sign chart for $f^{''} (x)$ :

Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on the polynomial factor $(4 x^{2} - 2)$ .

Interval	Test point $x$	Sign of $f^{''} (x)$
$(- \infty, - \frac{2}{2})$	$- 1$	$+$
$(- \frac{2}{2}, \frac{2}{2})$	$0$	$-$
$(\frac{2}{2}, \infty)$	$1$	$+$

Conclusion:

The sign of $f^{''} (x)$ changes at both $x = - \frac{2}{2}$ and $x = \frac{2}{2}$ . Therefore, $f$ has points of inflection at both $x$ -coordinates.

Intervals of concavity:

$f (x)$ is concave up for $x < - \frac{2}{2}$
Concave down for $- \frac{2}{2} < x < \frac{2}{2}$
Concave up for $x > \frac{2}{2}$

Example 2: Trigonometric/Interval Restricted

Let $f$ be the function given by $f (x) = x + 2 cos x$ . On the interval $[0, 2 π]$ , on which intervals is the graph of $f$ concave down?

Solution

(spoiler)

1. First derivative:

$f^{'} (x) = 1 - 2 sin x$

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ on the interval $[0, 2 π]$ to find potential inflection points:

$- 2 cos x = 0 cos x = 0 x = \frac{π}{2}, \frac{3 π}{2}$

4. Sign chart:

Analyze the sign of $f^{''} (x) = - 2 cos x$ across the subintervals:

Interval	Test Point $x$	Sign of $f^{''} (x)$
$[0, \frac{π}{2})$	$0$	$-$
$(\frac{π}{2}, \frac{3 π}{2})$	$π$	$+$
$(\frac{3 π}{2}, 2 π]$	$2 π$	$-$

Conclusion:

The graph of $f$ is concave down on the intervals $[0, \frac{π}{2})$ and $(\frac{3 π}{2}, 2 π]$ because $f^{''} (x) < 0$ on these intervals.

Example 3: No sign change

Find all $x$ -coordinates of the points of inflection for the function

$f (x) = \frac{3}{5} x^{5} - x^{4} + 1$

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

$f^{''} (x) = 12 x^{3} - 12 x^{2} = 12 x^{2} (x - 1)$

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$12 x^{2} (x - 1) = 0 x = 0, 1$

4. Sign chart:

Since $12 x^{2} > 0$ , the sign of $f^{''} (x)$ depends only on the factor $(x - 1)$ .

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, 0)$	$- 1$	$-$
$(0, 1)$	$\frac{1}{2}$	$-$
$(1, \infty)$	$2$	$+$

Conclusion: The sign of $f^{''} (x)$ changes from negative to positive at $x = 1$ . It does not change sign at $x = 0$ . Therefore, $x = 1$ is the only inflection point of $f$ .

Example 4: Working from the derivative

A function $f$ is continuous for all real numbers and its first derivative is given by $f^{'} (x) = (x - 2)^{2} (x + 3)$ . Find the $x$ -coordinate of each point of inflection of the graph of $f$ .

Solution

(spoiler)

To find where the concavity of $f$ changes, find $f^{''} (x)$ by differentiating $f^{'} (x)$ once:

$f^{''} (x) = \frac{d}{d x} [(x - 2)^{2}] \cdot (x + 3) + (x - 2)^{2} \cdot \frac{d}{d x} [x + 3] = 2 (x - 2) (x + 3) + (x - 2)^{2}$

Factor out the common term $(x - 2)$ :

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

Set $f^{''} (x) = 0$ to find potential inflection points:

$(x - 2) (3 x + 4) = 0 x = 2, - \frac{4}{3}$

Create a sign chart for $f^{''} (x)$ :

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, - \frac{4}{3})$	$- 2$	$+$
$(- \frac{4}{3}, 2)$	$0$	$-$
$(2, \infty)$	$3$	$+$

Conclusion: The graph of $f$ has points of inflection at $x = - \frac{4}{3}$ and $x = 2$ because $f^{''} (x)$ changes sign at these values.

Inflection points

Occur where a function changes concavity (up → down or down → up)
Potential inflection points: where f’'(x) = 0 or is undefined
Must confirm sign change in f’'(x) across the point — no sign change means no inflection point

Finding inflection points (steps)

Find f’'(x), set equal to zero (or find where undefined)
Build a sign chart using those x-values as boundaries
Inflection point confirmed only if f’'(x) changes sign across the boundary

Intervals of concavity

f’'(x) > 0 → concave up on that interval
f’'(x) < 0 → concave down on that interval

Key pitfalls

f’'(x) = 0 is necessary but not sufficient — always verify sign change
- e.g., f’'(x) = 12x²(x−1): x = 0 is not an inflection point (no sign change)
When given f’(x), differentiate once more to get f’'(x), then apply sign chart

Inflection points

What you’ll learn:

Points of inflection: Identify where a function changes concavity.
Intervals of concavity: Determine where a graph bends upward or downward based on the sign of the 2nd derivative.

An inflection point (or point of inflection) occurs when the graph of a function changes concavity, from concave up to concave down or vice versa.

A potential inflection point occurs where the second derivative $f^{''} (x) = 0$ or is undefined. To confirm it is an inflection point, the sign of $f^{''} (x)$ must change across that point.

Finding points of inflection:

Find all values of $x$ where $f^{''} (x) = 0$ or is undefined. These values must lie within the domain of the original function.
Create a sign chart for $f^{''} (x)$ using these values as interval boundaries.
If $f^{''} (x)$ changes sign across a boundary, that boundary point is an inflection point.

Example 1: Analytical

Find the points of inflection and determine the intervals of concavity for

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

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$e^{- x^{2}} (4 x^{2} - 2) = 0$

Because $e^{- x^{2}} > 0$ for all $x$ , solve:

$4 x^{2} - 2 x^{2} x = 0 = \frac{1}{2} = \pm \frac{2}{2}$

4. Sign chart for $f^{''} (x)$ :

Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on the polynomial factor $(4 x^{2} - 2)$ .

Interval	Test point $x$	Sign of $f^{''} (x)$
$(- \infty, - \frac{2}{2})$	$- 1$	$+$
$(- \frac{2}{2}, \frac{2}{2})$	$0$	$-$
$(\frac{2}{2}, \infty)$	$1$	$+$

Conclusion:

The sign of $f^{''} (x)$ changes at both $x = - \frac{2}{2}$ and $x = \frac{2}{2}$ . Therefore, $f$ has points of inflection at both $x$ -coordinates.

Intervals of concavity:

$f (x)$ is concave up for $x < - \frac{2}{2}$
Concave down for $- \frac{2}{2} < x < \frac{2}{2}$
Concave up for $x > \frac{2}{2}$

Example 2: Trigonometric/Interval Restricted

Let $f$ be the function given by $f (x) = x + 2 cos x$ . On the interval $[0, 2 π]$ , on which intervals is the graph of $f$ concave down?

Solution

(spoiler)

1. First derivative:

$f^{'} (x) = 1 - 2 sin x$

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ on the interval $[0, 2 π]$ to find potential inflection points:

$- 2 cos x = 0 cos x = 0 x = \frac{π}{2}, \frac{3 π}{2}$

4. Sign chart:

Analyze the sign of $f^{''} (x) = - 2 cos x$ across the subintervals:

Interval	Test Point $x$	Sign of $f^{''} (x)$
$[0, \frac{π}{2})$	$0$	$-$
$(\frac{π}{2}, \frac{3 π}{2})$	$π$	$+$
$(\frac{3 π}{2}, 2 π]$	$2 π$	$-$

Conclusion:

The graph of $f$ is concave down on the intervals $[0, \frac{π}{2})$ and $(\frac{3 π}{2}, 2 π]$ because $f^{''} (x) < 0$ on these intervals.

Example 3: No sign change

Find all $x$ -coordinates of the points of inflection for the function

$f (x) = \frac{3}{5} x^{5} - x^{4} + 1$

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

$f^{''} (x) = 12 x^{3} - 12 x^{2} = 12 x^{2} (x - 1)$

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$12 x^{2} (x - 1) = 0 x = 0, 1$

4. Sign chart:

Since $12 x^{2} > 0$ , the sign of $f^{''} (x)$ depends only on the factor $(x - 1)$ .

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, 0)$	$- 1$	$-$
$(0, 1)$	$\frac{1}{2}$	$-$
$(1, \infty)$	$2$	$+$

Conclusion: The sign of $f^{''} (x)$ changes from negative to positive at $x = 1$ . It does not change sign at $x = 0$ . Therefore, $x = 1$ is the only inflection point of $f$ .

Example 4: Working from the derivative

A function $f$ is continuous for all real numbers and its first derivative is given by $f^{'} (x) = (x - 2)^{2} (x + 3)$ . Find the $x$ -coordinate of each point of inflection of the graph of $f$ .

Solution

(spoiler)

To find where the concavity of $f$ changes, find $f^{''} (x)$ by differentiating $f^{'} (x)$ once:

$f^{''} (x) = \frac{d}{d x} [(x - 2)^{2}] \cdot (x + 3) + (x - 2)^{2} \cdot \frac{d}{d x} [x + 3] = 2 (x - 2) (x + 3) + (x - 2)^{2}$

Factor out the common term $(x - 2)$ :

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

Set $f^{''} (x) = 0$ to find potential inflection points:

$(x - 2) (3 x + 4) = 0 x = 2, - \frac{4}{3}$

Create a sign chart for $f^{''} (x)$ :

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, - \frac{4}{3})$	$- 2$	$+$
$(- \frac{4}{3}, 2)$	$0$	$-$
$(2, \infty)$	$3$	$+$

Conclusion: The graph of $f$ has points of inflection at $x = - \frac{4}{3}$ and $x = 2$ because $f^{''} (x)$ changes sign at these values.

Achievable AP Calculus AB

5. Analytical uses

5.3. 2nd derivative test

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Inflection points

5 min read

Font

Discuss

Feedback

What you’ll learn:

Points of inflection: Identify where a function changes concavity.
Intervals of concavity: Determine where a graph bends upward or downward based on the sign of the 2nd derivative.

An inflection point (or point of inflection) occurs when the graph of a function changes concavity, from concave up to concave down or vice versa.

A potential inflection point occurs where the second derivative $f^{''} (x) = 0$ or is undefined. To confirm it is an inflection point, the sign of $f^{''} (x)$ must change across that point.

Finding points of inflection:

Find all values of $x$ where $f^{''} (x) = 0$ or is undefined. These values must lie within the domain of the original function.
Create a sign chart for $f^{''} (x)$ using these values as interval boundaries.
If $f^{''} (x)$ changes sign across a boundary, that boundary point is an inflection point.

Example 1: Analytical

Find the points of inflection and determine the intervals of concavity for

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

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$e^{- x^{2}} (4 x^{2} - 2) = 0$

Because $e^{- x^{2}} > 0$ for all $x$ , solve:

$4 x^{2} - 2 x^{2} x = 0 = \frac{1}{2} = \pm \frac{2}{2}$

4. Sign chart for $f^{''} (x)$ :

Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on the polynomial factor $(4 x^{2} - 2)$ .

Interval	Test point $x$	Sign of $f^{''} (x)$
$(- \infty, - \frac{2}{2})$	$- 1$	$+$
$(- \frac{2}{2}, \frac{2}{2})$	$0$	$-$
$(\frac{2}{2}, \infty)$	$1$	$+$

Conclusion:

The sign of $f^{''} (x)$ changes at both $x = - \frac{2}{2}$ and $x = \frac{2}{2}$ . Therefore, $f$ has points of inflection at both $x$ -coordinates.

Intervals of concavity:

$f (x)$ is concave up for $x < - \frac{2}{2}$
Concave down for $- \frac{2}{2} < x < \frac{2}{2}$
Concave up for $x > \frac{2}{2}$

Example 2: Trigonometric/Interval Restricted

Let $f$ be the function given by $f (x) = x + 2 cos x$ . On the interval $[0, 2 π]$ , on which intervals is the graph of $f$ concave down?

Solution

(spoiler)

1. First derivative:

$f^{'} (x) = 1 - 2 sin x$

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ on the interval $[0, 2 π]$ to find potential inflection points:

$- 2 cos x = 0 cos x = 0 x = \frac{π}{2}, \frac{3 π}{2}$

4. Sign chart:

Analyze the sign of $f^{''} (x) = - 2 cos x$ across the subintervals:

Interval	Test Point $x$	Sign of $f^{''} (x)$
$[0, \frac{π}{2})$	$0$	$-$
$(\frac{π}{2}, \frac{3 π}{2})$	$π$	$+$
$(\frac{3 π}{2}, 2 π]$	$2 π$	$-$

Conclusion:

The graph of $f$ is concave down on the intervals $[0, \frac{π}{2})$ and $(\frac{3 π}{2}, 2 π]$ because $f^{''} (x) < 0$ on these intervals.

Example 3: No sign change

Find all $x$ -coordinates of the points of inflection for the function

$f (x) = \frac{3}{5} x^{5} - x^{4} + 1$

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

$f^{''} (x) = 12 x^{3} - 12 x^{2} = 12 x^{2} (x - 1)$

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$12 x^{2} (x - 1) = 0 x = 0, 1$

4. Sign chart:

Since $12 x^{2} > 0$ , the sign of $f^{''} (x)$ depends only on the factor $(x - 1)$ .

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, 0)$	$- 1$	$-$
$(0, 1)$	$\frac{1}{2}$	$-$
$(1, \infty)$	$2$	$+$

Conclusion: The sign of $f^{''} (x)$ changes from negative to positive at $x = 1$ . It does not change sign at $x = 0$ . Therefore, $x = 1$ is the only inflection point of $f$ .

Example 4: Working from the derivative

A function $f$ is continuous for all real numbers and its first derivative is given by $f^{'} (x) = (x - 2)^{2} (x + 3)$ . Find the $x$ -coordinate of each point of inflection of the graph of $f$ .

Solution

(spoiler)

To find where the concavity of $f$ changes, find $f^{''} (x)$ by differentiating $f^{'} (x)$ once:

$f^{''} (x) = \frac{d}{d x} [(x - 2)^{2}] \cdot (x + 3) + (x - 2)^{2} \cdot \frac{d}{d x} [x + 3] = 2 (x - 2) (x + 3) + (x - 2)^{2}$

Factor out the common term $(x - 2)$ :

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

Set $f^{''} (x) = 0$ to find potential inflection points:

$(x - 2) (3 x + 4) = 0 x = 2, - \frac{4}{3}$

Create a sign chart for $f^{''} (x)$ :

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, - \frac{4}{3})$	$- 2$	$+$
$(- \frac{4}{3}, 2)$	$0$	$-$
$(2, \infty)$	$3$	$+$

Conclusion: The graph of $f$ has points of inflection at $x = - \frac{4}{3}$ and $x = 2$ because $f^{''} (x)$ changes sign at these values.

Inflection points

Occur where a function changes concavity (up → down or down → up)
Potential inflection points: where f’'(x) = 0 or is undefined
Must confirm sign change in f’'(x) across the point — no sign change means no inflection point

Finding inflection points (steps)

Find f’'(x), set equal to zero (or find where undefined)
Build a sign chart using those x-values as boundaries
Inflection point confirmed only if f’'(x) changes sign across the boundary

Intervals of concavity

f’'(x) > 0 → concave up on that interval
f’'(x) < 0 → concave down on that interval

Key pitfalls

f’'(x) = 0 is necessary but not sufficient — always verify sign change
- e.g., f’'(x) = 12x²(x−1): x = 0 is not an inflection point (no sign change)
When given f’(x), differentiate once more to get f’'(x), then apply sign chart

Inflection points

What you’ll learn:

Points of inflection: Identify where a function changes concavity.
Intervals of concavity: Determine where a graph bends upward or downward based on the sign of the 2nd derivative.

An inflection point (or point of inflection) occurs when the graph of a function changes concavity, from concave up to concave down or vice versa.

A potential inflection point occurs where the second derivative $f^{''} (x) = 0$ or is undefined. To confirm it is an inflection point, the sign of $f^{''} (x)$ must change across that point.

Finding points of inflection:

Find all values of $x$ where $f^{''} (x) = 0$ or is undefined. These values must lie within the domain of the original function.
Create a sign chart for $f^{''} (x)$ using these values as interval boundaries.
If $f^{''} (x)$ changes sign across a boundary, that boundary point is an inflection point.

Example 1: Analytical

Find the points of inflection and determine the intervals of concavity for

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

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$e^{- x^{2}} (4 x^{2} - 2) = 0$

Because $e^{- x^{2}} > 0$ for all $x$ , solve:

$4 x^{2} - 2 x^{2} x = 0 = \frac{1}{2} = \pm \frac{2}{2}$

4. Sign chart for $f^{''} (x)$ :

Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on the polynomial factor $(4 x^{2} - 2)$ .

Interval	Test point $x$	Sign of $f^{''} (x)$
$(- \infty, - \frac{2}{2})$	$- 1$	$+$
$(- \frac{2}{2}, \frac{2}{2})$	$0$	$-$
$(\frac{2}{2}, \infty)$	$1$	$+$

Conclusion:

The sign of $f^{''} (x)$ changes at both $x = - \frac{2}{2}$ and $x = \frac{2}{2}$ . Therefore, $f$ has points of inflection at both $x$ -coordinates.

Intervals of concavity:

$f (x)$ is concave up for $x < - \frac{2}{2}$
Concave down for $- \frac{2}{2} < x < \frac{2}{2}$
Concave up for $x > \frac{2}{2}$

Example 2: Trigonometric/Interval Restricted

Let $f$ be the function given by $f (x) = x + 2 cos x$ . On the interval $[0, 2 π]$ , on which intervals is the graph of $f$ concave down?

Solution

(spoiler)

1. First derivative:

$f^{'} (x) = 1 - 2 sin x$

2. Second derivative:

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

3. Potential inflection points:

Set $f^{''} (x) = 0$ on the interval $[0, 2 π]$ to find potential inflection points:

$- 2 cos x = 0 cos x = 0 x = \frac{π}{2}, \frac{3 π}{2}$

4. Sign chart:

Analyze the sign of $f^{''} (x) = - 2 cos x$ across the subintervals:

Interval	Test Point $x$	Sign of $f^{''} (x)$
$[0, \frac{π}{2})$	$0$	$-$
$(\frac{π}{2}, \frac{3 π}{2})$	$π$	$+$
$(\frac{3 π}{2}, 2 π]$	$2 π$	$-$

Conclusion:

The graph of $f$ is concave down on the intervals $[0, \frac{π}{2})$ and $(\frac{3 π}{2}, 2 π]$ because $f^{''} (x) < 0$ on these intervals.

Example 3: No sign change

Find all $x$ -coordinates of the points of inflection for the function

$f (x) = \frac{3}{5} x^{5} - x^{4} + 1$

Solution

(spoiler)

The domain of $f (x)$ is all real numbers.

1. First derivative:

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

2. Second derivative:

$f^{''} (x) = 12 x^{3} - 12 x^{2} = 12 x^{2} (x - 1)$

3. Potential inflection points:

Set $f^{''} (x) = 0$ :

$12 x^{2} (x - 1) = 0 x = 0, 1$

4. Sign chart:

Since $12 x^{2} > 0$ , the sign of $f^{''} (x)$ depends only on the factor $(x - 1)$ .

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, 0)$	$- 1$	$-$
$(0, 1)$	$\frac{1}{2}$	$-$
$(1, \infty)$	$2$	$+$

Conclusion: The sign of $f^{''} (x)$ changes from negative to positive at $x = 1$ . It does not change sign at $x = 0$ . Therefore, $x = 1$ is the only inflection point of $f$ .

Example 4: Working from the derivative

A function $f$ is continuous for all real numbers and its first derivative is given by $f^{'} (x) = (x - 2)^{2} (x + 3)$ . Find the $x$ -coordinate of each point of inflection of the graph of $f$ .

Solution

(spoiler)

To find where the concavity of $f$ changes, find $f^{''} (x)$ by differentiating $f^{'} (x)$ once:

$f^{''} (x) = \frac{d}{d x} [(x - 2)^{2}] \cdot (x + 3) + (x - 2)^{2} \cdot \frac{d}{d x} [x + 3] = 2 (x - 2) (x + 3) + (x - 2)^{2}$

Factor out the common term $(x - 2)$ :

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

Set $f^{''} (x) = 0$ to find potential inflection points:

$(x - 2) (3 x + 4) = 0 x = 2, - \frac{4}{3}$

Create a sign chart for $f^{''} (x)$ :

Interval	Test Point	Sign of $f^{''} (x)$
$(- \infty, - \frac{4}{3})$	$- 2$	$+$
$(- \frac{4}{3}, 2)$	$0$	$-$
$(2, \infty)$	$3$	$+$

Conclusion: The graph of $f$ has points of inflection at $x = - \frac{4}{3}$ and $x = 2$ because $f^{''} (x)$ changes sign at these values.