5.3.2 Inflection points

Achievable AP Calculus AB

5. Analytical uses

5.3. 2nd derivative test

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Inflection points

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What you’ll learn:

How to find points of inflection/inflection points
Determine intervals of concavity 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 point of inflection happens where the 2nd derivative $f^{''} (x) = 0$ or is undefined. But the sign of $f^{''} (x)$ must change around the potential point for it to be an inflection point. Similar to how we test for local extrema using the sign changes of $f^{'} (x)$ , a sign diagram or chart for $f^{''} (x)$ must be used to confirm the point of inflection.

To find points of inflection:

Find where $f^{''} (x) = 0$ or undefined. These potential inflection points must also be in the domain of the original function.
Make a sign chart/diagram for $f^{''} (x)$ to check whether the sign changes across those points.
If the sign changes over a point, then it’s an inflection point.

Examples

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

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

The 1st derivative is

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

The 2nd derivative is

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

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

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

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

Since $e^{- x^{2}} > 0$ for all $x$ , it never equals $0$ . So we solve:

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

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

$x = \pm \frac{2}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on $4 x^{2} - 2$ .

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

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = - \frac{2}{2}$ as well as $x = \frac{2}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave up on $(\infty, - \frac{2}{2})$
Concave down on $(- \frac{2}{2}, \frac{2}{2})$
Concave up on $(\frac{2}{2}, \infty)$

Find the points of inflection and determine the concavity of

$g (x) = x^{2} - 1$

Solution

(spoiler)

For the domain of $g (x)$ , the expression under the square root must be non-negative.

$x^{2} - 1 \geq 0$

$(x + 1) (x - 1) \geq 0$

Use a number line and separate it at the $x$ -intercepts of $x = - 1$ and $x = 1$ to find that the expression is $\geq 0$ when $x \leq - 1$ or $x \geq 1$ .

Next, the 1st derivative is

$g^{'} (x) = \frac{1}{2} (x^{2} - 1)^{- 1/2} (2 x)$

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

The 2nd derivative is

$g^{''} (x) = \frac{( x ^{2} - 1 ) ^{1/2} - x \cdot \frac{x}{x ^{2} - 1}}{x ^{2} - 1}$

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

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

$g^{''} (x)$ is undefined when $x = 1$ and $- 1$ . However, $g (x)$ doesn’t exist in between $- 1$ and $1$ . Since we can’t check for a change in concavity, those aren’t inflection points.

The denominator is always positive, so $g^{''} (x) < 0$ for all $x$ in the domain of $g (x)$ . Therefore $g (x)$ is always concave down where it exists.

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

The domain is where the log argument $(x^{2} - x + 1) > 0$ . Checking the discriminant, $x^{2} - x + 1$ has no roots and is always above the $x$ -axis, so the domain of $f (x)$ is all real numbers.

The 1st derivative is

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

The 2nd derivative is

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

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

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

Since the denominator is always positive, $f^{''} (x)$ is never undefined. Solving $f^{''} (x) = 0$ for potential inflection points,

$- 2 x^{2} + 2 x + 1 = 0$

Using the quadratic formula,

$x = \frac{1 \pm 3}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $(x^{2} - x + 1)^{2} > 0$ , the sign of $f^{''} (x)$ depends only on $- 2 x^{2} + 2 x + 1$ .

Interval	Test point $x =$	Sign of $f^{''} (x)$
$(- \infty, \frac{1 - 3}{2})$	$- 1$	$-$
$(\frac{1 - 3}{2}, \frac{1 + 3}{2})$	$0$	$+$
$(\frac{1 + 3}{2}, \infty)$	$2$	$-$

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = \frac{1 - 3}{2}$ as well as $x = \frac{1 + 3}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave down on $(- \infty, \frac{1 - 3}{2})$
Concave up on $(\frac{1 - 3}{2}, \frac{1 + 3}{2})$
Concave down on $(\frac{1 + 3}{2}, \infty)$

A function $f$ has 1st derivative

$f^{'} (x) = x^{2} - 4 x - 5$

Find the points of inflection and determine the concavity of $f (x)$ .

Solution

(spoiler)

The 1st derivative is the given polynomial. The domain is all real numbers because only polynomials have derivatives that are also polynomials. The 2nd derivative is

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

$f^{''} (x) = 0$ when $x = 2$ , which is the potential inflection point.

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

Inflection points:

$x = 2$ is an inflection point since the sign of $g^{''} (x)$ changes.

Concavity:

$f (x)$ is concave down on $(- \infty, 2)$
Concave up on $(2, \infty)$

Inflection points and concavity

Inflection point: where function changes concavity (concave up ↔ concave down)
Potential inflection points: where $f^{''} (x) = 0$ or $f^{''} (x)$ undefined, and in domain of $f (x)$
Confirm inflection point: sign of $f^{''} (x)$ must change across the point

Finding inflection points (general steps)

Find where $f^{''} (x) = 0$ or $f^{''} (x)$ undefined, within domain
Use sign chart for $f^{''} (x)$ to check for sign change
Inflection point exists if sign of $f^{''} (x)$ changes

Example 1: $f (x) = e^{- x^{2}}$

Inflection points: $x = \pm \frac{2}{2}$
Concave up: $(- \infty, - \frac{2}{2})$ , $(\frac{2}{2}, \infty)$
Concave down: $(- \frac{2}{2}, \frac{2}{2})$

Example 2: $g (x) = x^{2} - 1$

Domain: $x \leq - 1$ or $x \geq 1$
$g^{''} (x)$ negative everywhere in domain
Always concave down; no inflection points

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

Inflection points: $x = \frac{1 \pm 3}{2}$
Concave down: $(- \infty, \frac{1 - 3}{2})$ , $(\frac{1 + 3}{2}, \infty)$
Concave up: $(\frac{1 - 3}{2}, \frac{1 + 3}{2})$

Example 4: $f^{'} (x) = x^{2} - 4 x - 5$

2nd derivative: $f^{''} (x) = 2 x - 4$
Inflection point: $x = 2$
Concave down: $(- \infty, 2)$
Concave up: $(2, \infty)$

Inflection points

What you’ll learn:

How to find points of inflection/inflection points
Determine intervals of concavity 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.

To find points of inflection:

Find where $f^{''} (x) = 0$ or undefined. These potential inflection points must also be in the domain of the original function.
Make a sign chart/diagram for $f^{''} (x)$ to check whether the sign changes across those points.
If the sign changes over a point, then it’s an inflection point.

Examples

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

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

The 1st derivative is

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

The 2nd derivative is

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

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

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

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

Since $e^{- x^{2}} > 0$ for all $x$ , it never equals $0$ . So we solve:

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

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

$x = \pm \frac{2}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on $4 x^{2} - 2$ .

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

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = - \frac{2}{2}$ as well as $x = \frac{2}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave up on $(\infty, - \frac{2}{2})$
Concave down on $(- \frac{2}{2}, \frac{2}{2})$
Concave up on $(\frac{2}{2}, \infty)$

Find the points of inflection and determine the concavity of

$g (x) = x^{2} - 1$

Solution

(spoiler)

For the domain of $g (x)$ , the expression under the square root must be non-negative.

$x^{2} - 1 \geq 0$

$(x + 1) (x - 1) \geq 0$

Use a number line and separate it at the $x$ -intercepts of $x = - 1$ and $x = 1$ to find that the expression is $\geq 0$ when $x \leq - 1$ or $x \geq 1$ .

Next, the 1st derivative is

$g^{'} (x) = \frac{1}{2} (x^{2} - 1)^{- 1/2} (2 x)$

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

The 2nd derivative is

$g^{''} (x) = \frac{( x ^{2} - 1 ) ^{1/2} - x \cdot \frac{x}{x ^{2} - 1}}{x ^{2} - 1}$

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

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

$g^{''} (x)$ is undefined when $x = 1$ and $- 1$ . However, $g (x)$ doesn’t exist in between $- 1$ and $1$ . Since we can’t check for a change in concavity, those aren’t inflection points.

The denominator is always positive, so $g^{''} (x) < 0$ for all $x$ in the domain of $g (x)$ . Therefore $g (x)$ is always concave down where it exists.

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

The domain is where the log argument $(x^{2} - x + 1) > 0$ . Checking the discriminant, $x^{2} - x + 1$ has no roots and is always above the $x$ -axis, so the domain of $f (x)$ is all real numbers.

The 1st derivative is

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

The 2nd derivative is

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

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

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

Since the denominator is always positive, $f^{''} (x)$ is never undefined. Solving $f^{''} (x) = 0$ for potential inflection points,

$- 2 x^{2} + 2 x + 1 = 0$

Using the quadratic formula,

$x = \frac{1 \pm 3}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $(x^{2} - x + 1)^{2} > 0$ , the sign of $f^{''} (x)$ depends only on $- 2 x^{2} + 2 x + 1$ .

Interval	Test point $x =$	Sign of $f^{''} (x)$
$(- \infty, \frac{1 - 3}{2})$	$- 1$	$-$
$(\frac{1 - 3}{2}, \frac{1 + 3}{2})$	$0$	$+$
$(\frac{1 + 3}{2}, \infty)$	$2$	$-$

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = \frac{1 - 3}{2}$ as well as $x = \frac{1 + 3}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave down on $(- \infty, \frac{1 - 3}{2})$
Concave up on $(\frac{1 - 3}{2}, \frac{1 + 3}{2})$
Concave down on $(\frac{1 + 3}{2}, \infty)$

A function $f$ has 1st derivative

$f^{'} (x) = x^{2} - 4 x - 5$

Find the points of inflection and determine the concavity of $f (x)$ .

Solution

(spoiler)

The 1st derivative is the given polynomial. The domain is all real numbers because only polynomials have derivatives that are also polynomials. The 2nd derivative is

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

$f^{''} (x) = 0$ when $x = 2$ , which is the potential inflection point.

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

Inflection points:

$x = 2$ is an inflection point since the sign of $g^{''} (x)$ changes.

Concavity:

$f (x)$ is concave down on $(- \infty, 2)$
Concave up on $(2, \infty)$

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

6 min read

Font

Discuss

Feedback

What you’ll learn:

How to find points of inflection/inflection points
Determine intervals of concavity 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.

To find points of inflection:

Find where $f^{''} (x) = 0$ or undefined. These potential inflection points must also be in the domain of the original function.
Make a sign chart/diagram for $f^{''} (x)$ to check whether the sign changes across those points.
If the sign changes over a point, then it’s an inflection point.

Examples

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

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

The 1st derivative is

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

The 2nd derivative is

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

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

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

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

Since $e^{- x^{2}} > 0$ for all $x$ , it never equals $0$ . So we solve:

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

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

$x = \pm \frac{2}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on $4 x^{2} - 2$ .

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

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = - \frac{2}{2}$ as well as $x = \frac{2}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave up on $(\infty, - \frac{2}{2})$
Concave down on $(- \frac{2}{2}, \frac{2}{2})$
Concave up on $(\frac{2}{2}, \infty)$

Find the points of inflection and determine the concavity of

$g (x) = x^{2} - 1$

Solution

(spoiler)

For the domain of $g (x)$ , the expression under the square root must be non-negative.

$x^{2} - 1 \geq 0$

$(x + 1) (x - 1) \geq 0$

Use a number line and separate it at the $x$ -intercepts of $x = - 1$ and $x = 1$ to find that the expression is $\geq 0$ when $x \leq - 1$ or $x \geq 1$ .

Next, the 1st derivative is

$g^{'} (x) = \frac{1}{2} (x^{2} - 1)^{- 1/2} (2 x)$

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

The 2nd derivative is

$g^{''} (x) = \frac{( x ^{2} - 1 ) ^{1/2} - x \cdot \frac{x}{x ^{2} - 1}}{x ^{2} - 1}$

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

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

$g^{''} (x)$ is undefined when $x = 1$ and $- 1$ . However, $g (x)$ doesn’t exist in between $- 1$ and $1$ . Since we can’t check for a change in concavity, those aren’t inflection points.

The denominator is always positive, so $g^{''} (x) < 0$ for all $x$ in the domain of $g (x)$ . Therefore $g (x)$ is always concave down where it exists.

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

The domain is where the log argument $(x^{2} - x + 1) > 0$ . Checking the discriminant, $x^{2} - x + 1$ has no roots and is always above the $x$ -axis, so the domain of $f (x)$ is all real numbers.

The 1st derivative is

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

The 2nd derivative is

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

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

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

Since the denominator is always positive, $f^{''} (x)$ is never undefined. Solving $f^{''} (x) = 0$ for potential inflection points,

$- 2 x^{2} + 2 x + 1 = 0$

Using the quadratic formula,

$x = \frac{1 \pm 3}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $(x^{2} - x + 1)^{2} > 0$ , the sign of $f^{''} (x)$ depends only on $- 2 x^{2} + 2 x + 1$ .

Interval	Test point $x =$	Sign of $f^{''} (x)$
$(- \infty, \frac{1 - 3}{2})$	$- 1$	$-$
$(\frac{1 - 3}{2}, \frac{1 + 3}{2})$	$0$	$+$
$(\frac{1 + 3}{2}, \infty)$	$2$	$-$

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = \frac{1 - 3}{2}$ as well as $x = \frac{1 + 3}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave down on $(- \infty, \frac{1 - 3}{2})$
Concave up on $(\frac{1 - 3}{2}, \frac{1 + 3}{2})$
Concave down on $(\frac{1 + 3}{2}, \infty)$

A function $f$ has 1st derivative

$f^{'} (x) = x^{2} - 4 x - 5$

Find the points of inflection and determine the concavity of $f (x)$ .

Solution

(spoiler)

The 1st derivative is the given polynomial. The domain is all real numbers because only polynomials have derivatives that are also polynomials. The 2nd derivative is

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

$f^{''} (x) = 0$ when $x = 2$ , which is the potential inflection point.

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

Inflection points:

$x = 2$ is an inflection point since the sign of $g^{''} (x)$ changes.

Concavity:

$f (x)$ is concave down on $(- \infty, 2)$
Concave up on $(2, \infty)$

Inflection points and concavity

Inflection point: where function changes concavity (concave up ↔ concave down)
Potential inflection points: where $f^{''} (x) = 0$ or $f^{''} (x)$ undefined, and in domain of $f (x)$
Confirm inflection point: sign of $f^{''} (x)$ must change across the point

Finding inflection points (general steps)

Find where $f^{''} (x) = 0$ or $f^{''} (x)$ undefined, within domain
Use sign chart for $f^{''} (x)$ to check for sign change
Inflection point exists if sign of $f^{''} (x)$ changes

Example 1: $f (x) = e^{- x^{2}}$

Inflection points: $x = \pm \frac{2}{2}$
Concave up: $(- \infty, - \frac{2}{2})$ , $(\frac{2}{2}, \infty)$
Concave down: $(- \frac{2}{2}, \frac{2}{2})$

Example 2: $g (x) = x^{2} - 1$

Domain: $x \leq - 1$ or $x \geq 1$
$g^{''} (x)$ negative everywhere in domain
Always concave down; no inflection points

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

Inflection points: $x = \frac{1 \pm 3}{2}$
Concave down: $(- \infty, \frac{1 - 3}{2})$ , $(\frac{1 + 3}{2}, \infty)$
Concave up: $(\frac{1 - 3}{2}, \frac{1 + 3}{2})$

Example 4: $f^{'} (x) = x^{2} - 4 x - 5$

2nd derivative: $f^{''} (x) = 2 x - 4$
Inflection point: $x = 2$
Concave down: $(- \infty, 2)$
Concave up: $(2, \infty)$

Inflection points

What you’ll learn:

How to find points of inflection/inflection points
Determine intervals of concavity 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.

To find points of inflection:

Find where $f^{''} (x) = 0$ or undefined. These potential inflection points must also be in the domain of the original function.
Make a sign chart/diagram for $f^{''} (x)$ to check whether the sign changes across those points.
If the sign changes over a point, then it’s an inflection point.

Examples

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

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

The 1st derivative is

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

The 2nd derivative is

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

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

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

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

Since $e^{- x^{2}} > 0$ for all $x$ , it never equals $0$ . So we solve:

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

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

$x = \pm \frac{2}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $e^{- x^{2}} > 0$ , the sign of $f^{''} (x)$ depends only on $4 x^{2} - 2$ .

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

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = - \frac{2}{2}$ as well as $x = \frac{2}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave up on $(\infty, - \frac{2}{2})$
Concave down on $(- \frac{2}{2}, \frac{2}{2})$
Concave up on $(\frac{2}{2}, \infty)$

Find the points of inflection and determine the concavity of

$g (x) = x^{2} - 1$

Solution

(spoiler)

For the domain of $g (x)$ , the expression under the square root must be non-negative.

$x^{2} - 1 \geq 0$

$(x + 1) (x - 1) \geq 0$

Use a number line and separate it at the $x$ -intercepts of $x = - 1$ and $x = 1$ to find that the expression is $\geq 0$ when $x \leq - 1$ or $x \geq 1$ .

Next, the 1st derivative is

$g^{'} (x) = \frac{1}{2} (x^{2} - 1)^{- 1/2} (2 x)$

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

The 2nd derivative is

$g^{''} (x) = \frac{( x ^{2} - 1 ) ^{1/2} - x \cdot \frac{x}{x ^{2} - 1}}{x ^{2} - 1}$

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

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

$g^{''} (x)$ is undefined when $x = 1$ and $- 1$ . However, $g (x)$ doesn’t exist in between $- 1$ and $1$ . Since we can’t check for a change in concavity, those aren’t inflection points.

The denominator is always positive, so $g^{''} (x) < 0$ for all $x$ in the domain of $g (x)$ . Therefore $g (x)$ is always concave down where it exists.

Find the points of inflection and determine the concavity of

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

Solution

(spoiler)

The domain is where the log argument $(x^{2} - x + 1) > 0$ . Checking the discriminant, $x^{2} - x + 1$ has no roots and is always above the $x$ -axis, so the domain of $f (x)$ is all real numbers.

The 1st derivative is

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

The 2nd derivative is

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

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

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

Since the denominator is always positive, $f^{''} (x)$ is never undefined. Solving $f^{''} (x) = 0$ for potential inflection points,

$- 2 x^{2} + 2 x + 1 = 0$

Using the quadratic formula,

$x = \frac{1 \pm 3}{2}$

Now set up a sign chart for $f^{''} (x)$ . Since $(x^{2} - x + 1)^{2} > 0$ , the sign of $f^{''} (x)$ depends only on $- 2 x^{2} + 2 x + 1$ .

Interval	Test point $x =$	Sign of $f^{''} (x)$
$(- \infty, \frac{1 - 3}{2})$	$- 1$	$-$
$(\frac{1 - 3}{2}, \frac{1 + 3}{2})$	$0$	$+$
$(\frac{1 + 3}{2}, \infty)$	$2$	$-$

Inflection points:

Since the sign of $f^{''} (x)$ changes as the graph passes $x = \frac{1 - 3}{2}$ as well as $x = \frac{1 + 3}{2}$ , both are inflection points.

Concavity:

$f (x)$ is concave down on $(- \infty, \frac{1 - 3}{2})$
Concave up on $(\frac{1 - 3}{2}, \frac{1 + 3}{2})$
Concave down on $(\frac{1 + 3}{2}, \infty)$

A function $f$ has 1st derivative

$f^{'} (x) = x^{2} - 4 x - 5$

Find the points of inflection and determine the concavity of $f (x)$ .

Solution

(spoiler)

The 1st derivative is the given polynomial. The domain is all real numbers because only polynomials have derivatives that are also polynomials. The 2nd derivative is

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

$f^{''} (x) = 0$ when $x = 2$ , which is the potential inflection point.

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

Inflection points:

$x = 2$ is an inflection point since the sign of $g^{''} (x)$ changes.

Concavity:

$f (x)$ is concave down on $(- \infty, 2)$
Concave up on $(2, \infty)$