5.2.2 On critical points

Achievable AP Calculus AB

5. Analytical uses

5.2. 1st derivative test

Our AP Calculus AB course is currently in development and is a work-in-progress.

On critical points

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What you’ll learn:

A critical reminder about critical points
Use sign charts to analyze function behavior around discontinuities

In the previous section, we found critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined. However, $f^{'} (a)$ only matters if $f (a)$ is defined.

If $x = a$ is not in the domain of $f (x)$ , then $x = a$ cannot be a critical point, even if $f^{'} (a) = 0$ or $f^{'} (a)$ is undefined. A number is only a critical point if the function exists there.

In the following examples, potential critical points are eliminated because the original function is undefined at those values.

Classify the extrema on

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

Solution

1. Find critical points

The inside of this function was purposely chosen to match the rational function from example 2 on the previous page. Using the chain rule, the derivative is

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

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

Canceling the $(x^{2} + 1)$ on top and bottom and simplifying gives

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

From this form:

$f^{'} (x) = 0$ when $x = 1$ or $x = - 1$ .
$f^{'} (x)$ is undefined when $x = 0$ .

Now check the domain of the original function. Since $f (x) = ln (\frac{x}{x ^{2} + 1})$ , we need

$\frac{x}{x ^{2} + 1} > 0.$

Because $x^{2} + 1 > 0$ for all real $x$ , the sign is determined by the numerator $x$ , so the domain is

$x > 0.$

That means $f (x)$ (and therefore the derivative test) only applies for $x > 0$ .

$f (0)$ is not defined, so $x = 0$ is not a critical point.
$f (- 1)$ is not defined, so $x = - 1$ is not a critical point.

So the only critical point is $x = 1$ .

2. Sign chart

Because the domain of $f (x)$ is $x > 0$ , we only need to test intervals to the right of $0$ for the sign of $f^{'} (x)$ .

Interval	Test point $x =$	Sign of $f^{'} (x)$	$f (x)$ behavior
$(0, 1)$	$0.5$	$+$	$↗$
$(1, \infty)$	$2$	$-$	$↘$

3. Interpret

Around $x = 1$ , $f^{'} (x)$ changes from positive to negative. That means $f (x)$ increases and then decreases, so there is a relative maximum at $x = 1$ .

Next is another example where a potential critical point is eliminated because the function is discontinuous there.

Classify the extrema on

$f (x) = \frac{1}{x - 1}$

Solution

(spoiler)

1. Find critical points

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

Because the numerator is a constant, there is no value of $x$ where $f^{'} (x) = 0$ .

When $x = 1$ , $f^{'} (x)$ is undefined. But since $f (x)$ is also undefined at $x = 1$ , it is not a critical point.

So this function has no critical points. That’s fine - not every function has critical points.

Even without critical points, we can still use a sign chart to describe how the function behaves around the point of discontinuity. This idea will be important in section 5.4 on curve sketching.

2. Sign chart

Here we’ll reason directly from $f^{'} (x)$ instead of plugging in test points. The denominator $(x - 1)^{2}$ is always positive (except where it’s undefined), and the negative sign in front makes the whole derivative negative. So

$f^{'} (x) < 0 for all x \neq = 1.$

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 1)$	$-$	$↘$
$(1, \infty)$	$-$	$↘$

3. Interpret

The function is decreasing for all $x$ . There are no critical points, so there are no relative maxima or minima. The vertical asymptote at $x = 1$ means the function approaches $\infty$ and $- \infty$ on either side, so there is no absolute maximum or minimum either.

Even when a potential critical point is actually a discontinuity, it still belongs on the sign chart because it splits the domain into separate intervals. It won’t be a relative extremum, but the behavior of $f (x)$ might differ on either side. For example,

Classify the extrema on

$f (x) = x e^{1/ x}$

Solution

(spoiler)

1. Find critical points

The derivative is

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

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

The derivative $f^{'} (x)$ equals $0$ when either $e^{1/ x} = 0$ or $1 - \frac{1}{x} = 0$ .

$e^{1/ x}$ is always positive, so it never equals $0$

Solve $1 - \frac{1}{x} = 0$ :

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

$x = 1$

$f^{'} (x)$ is undefined when $x = 0$ . But $f (x)$ is also undefined at $x = 0$ , so $x = 0$ is not a critical point.

So $x = 1$ is the only critical point, and $x = 0$ is a point of discontinuity.

2. Sign chart

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 0)$	$+$	$↗$
$(0, 1)$	$-$	$↘$
$(1, \infty)$	$+$	$↗$

3. Interpret

At $x = 0$ , $f (x)$ is not continuous. Even though $f^{'} (x)$ changes from positive to negative across $0$ , that change does not create a relative maximum or minimum because the function isn’t defined there.

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive, so $f (x)$ decreases and then increases. That means there is a relative minimum at $x = 1$ .

On critical points

What you’ll learn:

A critical reminder about critical points
Use sign charts to analyze function behavior around discontinuities

In the previous section, we found critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined. However, $f^{'} (a)$ only matters if $f (a)$ is defined.

In the following examples, potential critical points are eliminated because the original function is undefined at those values.

Classify the extrema on

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

Solution

1. Find critical points

The inside of this function was purposely chosen to match the rational function from example 2 on the previous page. Using the chain rule, the derivative is

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

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

Canceling the $(x^{2} + 1)$ on top and bottom and simplifying gives

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

From this form:

$f^{'} (x) = 0$ when $x = 1$ or $x = - 1$ .
$f^{'} (x)$ is undefined when $x = 0$ .

Now check the domain of the original function. Since $f (x) = ln (\frac{x}{x ^{2} + 1})$ , we need

$\frac{x}{x ^{2} + 1} > 0.$

Because $x^{2} + 1 > 0$ for all real $x$ , the sign is determined by the numerator $x$ , so the domain is

$x > 0.$

That means $f (x)$ (and therefore the derivative test) only applies for $x > 0$ .

$f (0)$ is not defined, so $x = 0$ is not a critical point.
$f (- 1)$ is not defined, so $x = - 1$ is not a critical point.

So the only critical point is $x = 1$ .

2. Sign chart

Because the domain of $f (x)$ is $x > 0$ , we only need to test intervals to the right of $0$ for the sign of $f^{'} (x)$ .

Interval	Test point $x =$	Sign of $f^{'} (x)$	$f (x)$ behavior
$(0, 1)$	$0.5$	$+$	$↗$
$(1, \infty)$	$2$	$-$	$↘$

3. Interpret

Around $x = 1$ , $f^{'} (x)$ changes from positive to negative. That means $f (x)$ increases and then decreases, so there is a relative maximum at $x = 1$ .

Next is another example where a potential critical point is eliminated because the function is discontinuous there.

Classify the extrema on

$f (x) = \frac{1}{x - 1}$

Solution

(spoiler)

1. Find critical points

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

Because the numerator is a constant, there is no value of $x$ where $f^{'} (x) = 0$ .

When $x = 1$ , $f^{'} (x)$ is undefined. But since $f (x)$ is also undefined at $x = 1$ , it is not a critical point.

So this function has no critical points. That’s fine - not every function has critical points.

Even without critical points, we can still use a sign chart to describe how the function behaves around the point of discontinuity. This idea will be important in section 5.4 on curve sketching.

2. Sign chart

$f^{'} (x) < 0 for all x \neq = 1.$

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 1)$	$-$	$↘$
$(1, \infty)$	$-$	$↘$

3. Interpret

Classify the extrema on

$f (x) = x e^{1/ x}$

Solution

(spoiler)

1. Find critical points

The derivative is

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

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

The derivative $f^{'} (x)$ equals $0$ when either $e^{1/ x} = 0$ or $1 - \frac{1}{x} = 0$ .

$e^{1/ x}$ is always positive, so it never equals $0$

Solve $1 - \frac{1}{x} = 0$ :

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

$x = 1$

$f^{'} (x)$ is undefined when $x = 0$ . But $f (x)$ is also undefined at $x = 0$ , so $x = 0$ is not a critical point.

So $x = 1$ is the only critical point, and $x = 0$ is a point of discontinuity.

2. Sign chart

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 0)$	$+$	$↗$
$(0, 1)$	$-$	$↘$
$(1, \infty)$	$+$	$↗$

3. Interpret

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive, so $f (x)$ decreases and then increases. That means there is a relative minimum at $x = 1$ .

Achievable AP Calculus AB

5. Analytical uses

5.2. 1st derivative test

Our AP Calculus AB course is currently in development and is a work-in-progress.

On critical points

5 min read

Font

Discuss

Feedback

What you’ll learn:

A critical reminder about critical points
Use sign charts to analyze function behavior around discontinuities

In the previous section, we found critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined. However, $f^{'} (a)$ only matters if $f (a)$ is defined.

In the following examples, potential critical points are eliminated because the original function is undefined at those values.

Classify the extrema on

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

Solution

1. Find critical points

The inside of this function was purposely chosen to match the rational function from example 2 on the previous page. Using the chain rule, the derivative is

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

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

Canceling the $(x^{2} + 1)$ on top and bottom and simplifying gives

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

From this form:

$f^{'} (x) = 0$ when $x = 1$ or $x = - 1$ .
$f^{'} (x)$ is undefined when $x = 0$ .

Now check the domain of the original function. Since $f (x) = ln (\frac{x}{x ^{2} + 1})$ , we need

$\frac{x}{x ^{2} + 1} > 0.$

Because $x^{2} + 1 > 0$ for all real $x$ , the sign is determined by the numerator $x$ , so the domain is

$x > 0.$

That means $f (x)$ (and therefore the derivative test) only applies for $x > 0$ .

$f (0)$ is not defined, so $x = 0$ is not a critical point.
$f (- 1)$ is not defined, so $x = - 1$ is not a critical point.

So the only critical point is $x = 1$ .

2. Sign chart

Because the domain of $f (x)$ is $x > 0$ , we only need to test intervals to the right of $0$ for the sign of $f^{'} (x)$ .

Interval	Test point $x =$	Sign of $f^{'} (x)$	$f (x)$ behavior
$(0, 1)$	$0.5$	$+$	$↗$
$(1, \infty)$	$2$	$-$	$↘$

3. Interpret

Around $x = 1$ , $f^{'} (x)$ changes from positive to negative. That means $f (x)$ increases and then decreases, so there is a relative maximum at $x = 1$ .

Next is another example where a potential critical point is eliminated because the function is discontinuous there.

Classify the extrema on

$f (x) = \frac{1}{x - 1}$

Solution

(spoiler)

1. Find critical points

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

Because the numerator is a constant, there is no value of $x$ where $f^{'} (x) = 0$ .

When $x = 1$ , $f^{'} (x)$ is undefined. But since $f (x)$ is also undefined at $x = 1$ , it is not a critical point.

So this function has no critical points. That’s fine - not every function has critical points.

Even without critical points, we can still use a sign chart to describe how the function behaves around the point of discontinuity. This idea will be important in section 5.4 on curve sketching.

2. Sign chart

$f^{'} (x) < 0 for all x \neq = 1.$

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 1)$	$-$	$↘$
$(1, \infty)$	$-$	$↘$

3. Interpret

Classify the extrema on

$f (x) = x e^{1/ x}$

Solution

(spoiler)

1. Find critical points

The derivative is

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

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

The derivative $f^{'} (x)$ equals $0$ when either $e^{1/ x} = 0$ or $1 - \frac{1}{x} = 0$ .

$e^{1/ x}$ is always positive, so it never equals $0$

Solve $1 - \frac{1}{x} = 0$ :

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

$x = 1$

$f^{'} (x)$ is undefined when $x = 0$ . But $f (x)$ is also undefined at $x = 0$ , so $x = 0$ is not a critical point.

So $x = 1$ is the only critical point, and $x = 0$ is a point of discontinuity.

2. Sign chart

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 0)$	$+$	$↗$
$(0, 1)$	$-$	$↘$
$(1, \infty)$	$+$	$↗$

3. Interpret

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive, so $f (x)$ decreases and then increases. That means there is a relative minimum at $x = 1$ .

On critical points

What you’ll learn:

A critical reminder about critical points
Use sign charts to analyze function behavior around discontinuities

In the previous section, we found critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined. However, $f^{'} (a)$ only matters if $f (a)$ is defined.

In the following examples, potential critical points are eliminated because the original function is undefined at those values.

Classify the extrema on

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

Solution

1. Find critical points

The inside of this function was purposely chosen to match the rational function from example 2 on the previous page. Using the chain rule, the derivative is

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

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

Canceling the $(x^{2} + 1)$ on top and bottom and simplifying gives

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

From this form:

$f^{'} (x) = 0$ when $x = 1$ or $x = - 1$ .
$f^{'} (x)$ is undefined when $x = 0$ .

Now check the domain of the original function. Since $f (x) = ln (\frac{x}{x ^{2} + 1})$ , we need

$\frac{x}{x ^{2} + 1} > 0.$

Because $x^{2} + 1 > 0$ for all real $x$ , the sign is determined by the numerator $x$ , so the domain is

$x > 0.$

That means $f (x)$ (and therefore the derivative test) only applies for $x > 0$ .

$f (0)$ is not defined, so $x = 0$ is not a critical point.
$f (- 1)$ is not defined, so $x = - 1$ is not a critical point.

So the only critical point is $x = 1$ .

2. Sign chart

Because the domain of $f (x)$ is $x > 0$ , we only need to test intervals to the right of $0$ for the sign of $f^{'} (x)$ .

Interval	Test point $x =$	Sign of $f^{'} (x)$	$f (x)$ behavior
$(0, 1)$	$0.5$	$+$	$↗$
$(1, \infty)$	$2$	$-$	$↘$

3. Interpret

Around $x = 1$ , $f^{'} (x)$ changes from positive to negative. That means $f (x)$ increases and then decreases, so there is a relative maximum at $x = 1$ .

Next is another example where a potential critical point is eliminated because the function is discontinuous there.

Classify the extrema on

$f (x) = \frac{1}{x - 1}$

Solution

(spoiler)

1. Find critical points

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

Because the numerator is a constant, there is no value of $x$ where $f^{'} (x) = 0$ .

When $x = 1$ , $f^{'} (x)$ is undefined. But since $f (x)$ is also undefined at $x = 1$ , it is not a critical point.

So this function has no critical points. That’s fine - not every function has critical points.

Even without critical points, we can still use a sign chart to describe how the function behaves around the point of discontinuity. This idea will be important in section 5.4 on curve sketching.

2. Sign chart

$f^{'} (x) < 0 for all x \neq = 1.$

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 1)$	$-$	$↘$
$(1, \infty)$	$-$	$↘$

3. Interpret

Classify the extrema on

$f (x) = x e^{1/ x}$

Solution

(spoiler)

1. Find critical points

The derivative is

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

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

The derivative $f^{'} (x)$ equals $0$ when either $e^{1/ x} = 0$ or $1 - \frac{1}{x} = 0$ .

$e^{1/ x}$ is always positive, so it never equals $0$

Solve $1 - \frac{1}{x} = 0$ :

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

$x = 1$

$f^{'} (x)$ is undefined when $x = 0$ . But $f (x)$ is also undefined at $x = 0$ , so $x = 0$ is not a critical point.

So $x = 1$ is the only critical point, and $x = 0$ is a point of discontinuity.

2. Sign chart

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \infty, 0)$	$+$	$↗$
$(0, 1)$	$-$	$↘$
$(1, \infty)$	$+$	$↗$

3. Interpret

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive, so $f (x)$ decreases and then increases. That means there is a relative minimum at $x = 1$ .