5.2.1 Extrema

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.

Extrema

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What you’ll learn:

How to find critical points of various functions
How to use the 1st derivative test (and sign charts) to identify relative/local extrema

In the previous section, you learned how to find the highest and lowest values of a function on a closed interval using the extreme value theorem. The method was to compare the function values at:

the endpoints of the interval, and
any critical points inside the interval.

The 1st derivative test gives you a different tool: it helps you classify critical points (anywhere in the domain) as relative/local maxima, relative/local minima, or neither.

Absolute vs. relative extrema

A global (absolute) extremum is the highest or lowest value a function attains over its entire domain.

For example, consider $f (x) = x^{2}$ .

The vertex $(0, 0)$ is the global/absolute minimum because it’s the lowest point on the entire domain of all real numbers. It’s also a relative/local minimum because, compared to nearby points, it’s still lower.

On the other hand, $x^{2}$ has no absolute maximum on its full domain, because as $x \to \pm \infty$ , $x^{2} \to \infty$ .

First derivative test

Here’s how to find and classify extrema using the 1st derivative test:

1. Find critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined.

Important: $x = a$ is only a critical point if it’s in the domain of $f (x)$ (so $f (a)$ must be defined). The next page will discuss this in more detail, but it’s worth keeping in mind now.

2. Create a sign diagram or chart for $f^{'} (x)$ using intervals around each critical point. Determine the sign of $f^{'} (x)$ before and after each critical point by testing a point in each interval.

3. Interpret each critical point

If $f^{'} (x)$ changes from positive to negative at a critical point, then $f (x)$ increases ( $↗$ ) and then decreases ( $↘$ ). The critical point is a local/relative maximum (a peak).
If $f^{'} (x)$ changes from negative to positive at a critical point, then $f (x)$ decreases ( $↘$ ) and then increases ( $↗$ ). The critical point is a local/relative minimum (a valley).
If there is no sign change, the critical point is not an extremum. For example, $y = x^{3}$ is flat at $(0, 0)$ , but the function keeps increasing through that point.

Examples

We’ll start with a basic polynomial.

Classify the extrema on

$g (x) = (x - 2)^{2} (x + 1)^{3}$

Solution

1. Find critical points

Using the product rule,

$g^{'} (x) = (x - 2)^{2} \cdot 3 (x + 1)^{2} + (x + 1)^{3} \cdot 2 (x - 2)$

$= (x - 2) (x + 1)^{2} [3 (x - 2) + 2 (x + 1)]$

$= (x - 2) (x + 1)^{2} (5 x - 4)$

Since $g^{'} (x)$ is a polynomial, it’s defined for all real $x$ . To find critical points, solve $g^{'} (x) = 0$ :

$(x - 2) (x + 1)^{2} (5 x - 4) = 0$

$x = 2, - 1, \frac{4}{5}$

2. Create a sign diagram or chart

Here it’ll be displayed as a sign chart, but a sign diagram with the intervals displayed left to right is usually easier to read (especially when you graph later).

Interval	Test point $x =$	Sign of $g^{'} (x)$	$g (x)$ behavior
$(\infty, - 1)$	$- 2$	$+$	$↗$
$(- 1, \frac{4}{5})$	$0$	$+$	$↗$
$(\frac{4}{5}, 2)$	$1$	$-$	$↘$
$(2, \infty)$	$3$	$+$	$↘$

3. Interpret

From $(\infty, - 1)$ , $g^{'} (x)$ is positive, so $g (x)$ is increasing.

From $(- 1, \frac{4}{5})$ , $g^{'} (x)$ is still positive, so $g (x)$ continues increasing. Since there’s no sign change at $x = - 1$ , there is no relative maximum or minimum at $x = - 1$ .

Around $x = \frac{4}{5}$ , $g^{'} (x)$ changes from positive to negative. So $g (x)$ increases and then decreases, which means there is a relative maximum at $x = \frac{4}{5}$ .

Around $x = 2$ , $g^{'} (x)$ changes from negative to positive. So $g (x)$ decreases and then increases, which means there is a relative minimum at $x = 2$ .

Because $g (x)$ is a polynomial with an odd degree, it has no absolute extrema. As $x \to \infty$ , $g (x) \to \infty$ , and as $x \to - \infty$ , $g (x) \to - \infty$ .

AP tip:

In addition to a sign chart for $f^{'} (x)$ , you must state what it implies about $f (x)$ (increasing or decreasing) to justify a relative max, min, or neither at a critical point. A sign chart alone won’t be enough.

It suffices to make a statement such as “there is a relative maximum at $x = 1$ because $f (x)$ is increasing and then decreasing” or “because $f^{'} (x)$ changes from positive to negative.”

Next, let’s work with a rational function.

Classify the extrema on

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

Solution

(spoiler)

1. Find critical points

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

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

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

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

A rational function is undefined where its denominator is zero. Here, the denominator of $f^{'} (x)$ is $(x^{2} + 1)^{2}$ , and since $(x^{2} + 1)^{2} > 0$ for all real $x$ , $f^{'} (x)$ is defined everywhere.

So the critical points come only from $f^{'} (x) = 0$ , which happens when the numerator is zero. The critical points are $x = 1, - 1$ .

2. Sign chart

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

3. Interpret

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

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

Next, we’ll look at a function where $f^{'} (a)$ is undefined.

Classify the extrema on

$f (x) = ∣ x^{2} - 1∣$

Solution

(spoiler)

Rewriting $f (x)$ as a piecewise function,

$f (x) = ⎩ ⎨ ⎧ x^{2} - 1 - (x^{2} - 1) x^{2} - 1 if x < - 1 if - 1 < x < 1 if x > 1$

This means the derivative is

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 2 x 2 x if x < - 1 if - 1 < x < 1 if x > 1$

$f^{'} (x) = 0$ when $x = 0$ .

To find where $f^{'} (x)$ is undefined (where $f$ is not differentiable), focus on the breakpoints of the piecewise definition: $x = - 1$ and $x = 1$ . At a breakpoint, the derivative exists only if the left-hand derivative and right-hand derivative match.

For $x = - 1$ :

$Left:$

$x \to - 1^{-} lim f (x) = 2 (- 1) = - 2$

$Right:$

$x \to - 1^{+} lim f (x) = - 2 (- 1) = 2$

The two one-sided values don’t agree, so the derivative doesn’t exist at $x = - 1$ (so $f^{'} (x)$ is undefined there). But $f (- 1)$ exists (it equals 0), so $x = - 1$ is a critical point.

Similarly, for $x = 1$ ,

$Left: x \to 1^{-} lim f (x) = - 2 (1) = - 2$

$Right: x \to 1^{+} lim f (x) = 2 (1) = 2$

Since the two sides don’t agree, $f^{'} (1)$ is undefined and $x = 1$ is also a critical point.

We have three critical points: $x = - 1, 0, 1$ , so there are 4 intervals where we can test the sign of $f^{'} (x)$ .

2. Sign chart

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

3. Interpret

Around $x = - 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Around $x = 0$ , $f^{'} (x)$ changes from positive to negative $\to$ relative maximum.

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Lastly, we’ll discuss critical points that occur periodically. Trig functions often behave this way.

Classify all extrema on

$f (x) = e^{s i n (x)}$

Solution

(spoiler)

1. Find critical points

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

To solve $f^{'} (x) = 0$ , we need

$e^{s i n (x)} = 0$ or
$cos (x) = 0$ .

Since $e^{anything} > 0$ for all real inputs, $e^{s i n (x)}$ is never 0. So the only way for $f^{'} (x) = 0$ is $cos (x) = 0$ .

$cos (x) = 0$ when $x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, etc.$ These critical points repeat every $π$ .

2. Sign chart

Although $f (x)$ repeats forever, we’ll limit the sign chart to these intervals: $(- \frac{π}{2}, \frac{π}{2}), (\frac{π}{2}, \frac{3 π}{2}), (\frac{3 π}{2}, \frac{5 π}{2})$ .

Instead of testing points, we can determine the sign of $f^{'} (x)$ by reasoning:

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

$e^{s i n (x)}$ is always positive.
So the sign of $f^{'} (x)$ is the same as the sign of $cos (x)$ .

On $(- \frac{π}{2}, \frac{π}{2})$ , $cos (x) > 0$ (quadrants I and IV), so $f^{'} (x) > 0$ .

On $(\frac{π}{2}, \frac{3 π}{2})$ , $cos (x) < 0$ (quadrants II and III), so $f^{'} (x) < 0$ .

On $(\frac{3 π}{2}, \frac{5 π}{2})$ , $cos (x) > 0$ again, so $f^{'} (x) > 0$ .

Here is the organized sign chart:

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \frac{π}{2}, \frac{π}{2})$	$+$	$↗$
$(\frac{π}{2}, \frac{3 π}{2})$	$-$	$↘$
$(\frac{3 π}{2}, \frac{5 π}{2})$	$+$	$↗$

3. Interpret

Around $x = \frac{π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

Around $x = \frac{3 π}{2}$ , $f (x)$ decreases and then increases $\to$ relative min.

Around $x = \frac{5 π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

The critical points will continue to alternate between relative maxima and minima.

Extrema

What you’ll learn:

How to find critical points of various functions
How to use the 1st derivative test (and sign charts) to identify relative/local extrema

the endpoints of the interval, and
any critical points inside the interval.

The 1st derivative test gives you a different tool: it helps you classify critical points (anywhere in the domain) as relative/local maxima, relative/local minima, or neither.

Absolute vs. relative extrema

A global (absolute) extremum is the highest or lowest value a function attains over its entire domain.

For example, consider $f (x) = x^{2}$ .

On the other hand, $x^{2}$ has no absolute maximum on its full domain, because as $x \to \pm \infty$ , $x^{2} \to \infty$ .

First derivative test

Here’s how to find and classify extrema using the 1st derivative test:

1. Find critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined.

Important: $x = a$ is only a critical point if it’s in the domain of $f (x)$ (so $f (a)$ must be defined). The next page will discuss this in more detail, but it’s worth keeping in mind now.

3. Interpret each critical point

If $f^{'} (x)$ changes from positive to negative at a critical point, then $f (x)$ increases ( $↗$ ) and then decreases ( $↘$ ). The critical point is a local/relative maximum (a peak).
If $f^{'} (x)$ changes from negative to positive at a critical point, then $f (x)$ decreases ( $↘$ ) and then increases ( $↗$ ). The critical point is a local/relative minimum (a valley).
If there is no sign change, the critical point is not an extremum. For example, $y = x^{3}$ is flat at $(0, 0)$ , but the function keeps increasing through that point.

Examples

We’ll start with a basic polynomial.

Classify the extrema on

$g (x) = (x - 2)^{2} (x + 1)^{3}$

Solution

1. Find critical points

Using the product rule,

$g^{'} (x) = (x - 2)^{2} \cdot 3 (x + 1)^{2} + (x + 1)^{3} \cdot 2 (x - 2)$

$= (x - 2) (x + 1)^{2} [3 (x - 2) + 2 (x + 1)]$

$= (x - 2) (x + 1)^{2} (5 x - 4)$

Since $g^{'} (x)$ is a polynomial, it’s defined for all real $x$ . To find critical points, solve $g^{'} (x) = 0$ :

$(x - 2) (x + 1)^{2} (5 x - 4) = 0$

$x = 2, - 1, \frac{4}{5}$

2. Create a sign diagram or chart

Here it’ll be displayed as a sign chart, but a sign diagram with the intervals displayed left to right is usually easier to read (especially when you graph later).

Interval	Test point $x =$	Sign of $g^{'} (x)$	$g (x)$ behavior
$(\infty, - 1)$	$- 2$	$+$	$↗$
$(- 1, \frac{4}{5})$	$0$	$+$	$↗$
$(\frac{4}{5}, 2)$	$1$	$-$	$↘$
$(2, \infty)$	$3$	$+$	$↘$

3. Interpret

From $(\infty, - 1)$ , $g^{'} (x)$ is positive, so $g (x)$ is increasing.

From $(- 1, \frac{4}{5})$ , $g^{'} (x)$ is still positive, so $g (x)$ continues increasing. Since there’s no sign change at $x = - 1$ , there is no relative maximum or minimum at $x = - 1$ .

Around $x = \frac{4}{5}$ , $g^{'} (x)$ changes from positive to negative. So $g (x)$ increases and then decreases, which means there is a relative maximum at $x = \frac{4}{5}$ .

Around $x = 2$ , $g^{'} (x)$ changes from negative to positive. So $g (x)$ decreases and then increases, which means there is a relative minimum at $x = 2$ .

Because $g (x)$ is a polynomial with an odd degree, it has no absolute extrema. As $x \to \infty$ , $g (x) \to \infty$ , and as $x \to - \infty$ , $g (x) \to - \infty$ .

AP tip:

It suffices to make a statement such as “there is a relative maximum at $x = 1$ because $f (x)$ is increasing and then decreasing” or “because $f^{'} (x)$ changes from positive to negative.”

Next, let’s work with a rational function.

Classify the extrema on

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

Solution

(spoiler)

1. Find critical points

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

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

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

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

So the critical points come only from $f^{'} (x) = 0$ , which happens when the numerator is zero. The critical points are $x = 1, - 1$ .

2. Sign chart

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

3. Interpret

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

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

Next, we’ll look at a function where $f^{'} (a)$ is undefined.

Classify the extrema on

$f (x) = ∣ x^{2} - 1∣$

Solution

(spoiler)

Rewriting $f (x)$ as a piecewise function,

$f (x) = ⎩ ⎨ ⎧ x^{2} - 1 - (x^{2} - 1) x^{2} - 1 if x < - 1 if - 1 < x < 1 if x > 1$

This means the derivative is

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 2 x 2 x if x < - 1 if - 1 < x < 1 if x > 1$

$f^{'} (x) = 0$ when $x = 0$ .

For $x = - 1$ :

$Left:$

$x \to - 1^{-} lim f (x) = 2 (- 1) = - 2$

$Right:$

$x \to - 1^{+} lim f (x) = - 2 (- 1) = 2$

The two one-sided values don’t agree, so the derivative doesn’t exist at $x = - 1$ (so $f^{'} (x)$ is undefined there). But $f (- 1)$ exists (it equals 0), so $x = - 1$ is a critical point.

Similarly, for $x = 1$ ,

$Left: x \to 1^{-} lim f (x) = - 2 (1) = - 2$

$Right: x \to 1^{+} lim f (x) = 2 (1) = 2$

Since the two sides don’t agree, $f^{'} (1)$ is undefined and $x = 1$ is also a critical point.

We have three critical points: $x = - 1, 0, 1$ , so there are 4 intervals where we can test the sign of $f^{'} (x)$ .

2. Sign chart

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

3. Interpret

Around $x = - 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Around $x = 0$ , $f^{'} (x)$ changes from positive to negative $\to$ relative maximum.

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Lastly, we’ll discuss critical points that occur periodically. Trig functions often behave this way.

Classify all extrema on

$f (x) = e^{s i n (x)}$

Solution

(spoiler)

1. Find critical points

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

To solve $f^{'} (x) = 0$ , we need

$e^{s i n (x)} = 0$ or
$cos (x) = 0$ .

Since $e^{anything} > 0$ for all real inputs, $e^{s i n (x)}$ is never 0. So the only way for $f^{'} (x) = 0$ is $cos (x) = 0$ .

$cos (x) = 0$ when $x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, etc.$ These critical points repeat every $π$ .

2. Sign chart

Although $f (x)$ repeats forever, we’ll limit the sign chart to these intervals: $(- \frac{π}{2}, \frac{π}{2}), (\frac{π}{2}, \frac{3 π}{2}), (\frac{3 π}{2}, \frac{5 π}{2})$ .

Instead of testing points, we can determine the sign of $f^{'} (x)$ by reasoning:

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

$e^{s i n (x)}$ is always positive.
So the sign of $f^{'} (x)$ is the same as the sign of $cos (x)$ .

On $(- \frac{π}{2}, \frac{π}{2})$ , $cos (x) > 0$ (quadrants I and IV), so $f^{'} (x) > 0$ .

On $(\frac{π}{2}, \frac{3 π}{2})$ , $cos (x) < 0$ (quadrants II and III), so $f^{'} (x) < 0$ .

On $(\frac{3 π}{2}, \frac{5 π}{2})$ , $cos (x) > 0$ again, so $f^{'} (x) > 0$ .

Here is the organized sign chart:

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \frac{π}{2}, \frac{π}{2})$	$+$	$↗$
$(\frac{π}{2}, \frac{3 π}{2})$	$-$	$↘$
$(\frac{3 π}{2}, \frac{5 π}{2})$	$+$	$↗$

3. Interpret

Around $x = \frac{π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

Around $x = \frac{3 π}{2}$ , $f (x)$ decreases and then increases $\to$ relative min.

Around $x = \frac{5 π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

The critical points will continue to alternate between relative maxima and minima.

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.

Extrema

9 min read

Font

Discuss

Feedback

What you’ll learn:

How to find critical points of various functions
How to use the 1st derivative test (and sign charts) to identify relative/local extrema

the endpoints of the interval, and
any critical points inside the interval.

The 1st derivative test gives you a different tool: it helps you classify critical points (anywhere in the domain) as relative/local maxima, relative/local minima, or neither.

Absolute vs. relative extrema

A global (absolute) extremum is the highest or lowest value a function attains over its entire domain.

For example, consider $f (x) = x^{2}$ .

On the other hand, $x^{2}$ has no absolute maximum on its full domain, because as $x \to \pm \infty$ , $x^{2} \to \infty$ .

First derivative test

Here’s how to find and classify extrema using the 1st derivative test:

1. Find critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined.

Important: $x = a$ is only a critical point if it’s in the domain of $f (x)$ (so $f (a)$ must be defined). The next page will discuss this in more detail, but it’s worth keeping in mind now.

3. Interpret each critical point

If $f^{'} (x)$ changes from positive to negative at a critical point, then $f (x)$ increases ( $↗$ ) and then decreases ( $↘$ ). The critical point is a local/relative maximum (a peak).
If $f^{'} (x)$ changes from negative to positive at a critical point, then $f (x)$ decreases ( $↘$ ) and then increases ( $↗$ ). The critical point is a local/relative minimum (a valley).
If there is no sign change, the critical point is not an extremum. For example, $y = x^{3}$ is flat at $(0, 0)$ , but the function keeps increasing through that point.

Examples

We’ll start with a basic polynomial.

Classify the extrema on

$g (x) = (x - 2)^{2} (x + 1)^{3}$

Solution

1. Find critical points

Using the product rule,

$g^{'} (x) = (x - 2)^{2} \cdot 3 (x + 1)^{2} + (x + 1)^{3} \cdot 2 (x - 2)$

$= (x - 2) (x + 1)^{2} [3 (x - 2) + 2 (x + 1)]$

$= (x - 2) (x + 1)^{2} (5 x - 4)$

Since $g^{'} (x)$ is a polynomial, it’s defined for all real $x$ . To find critical points, solve $g^{'} (x) = 0$ :

$(x - 2) (x + 1)^{2} (5 x - 4) = 0$

$x = 2, - 1, \frac{4}{5}$

2. Create a sign diagram or chart

Here it’ll be displayed as a sign chart, but a sign diagram with the intervals displayed left to right is usually easier to read (especially when you graph later).

Interval	Test point $x =$	Sign of $g^{'} (x)$	$g (x)$ behavior
$(\infty, - 1)$	$- 2$	$+$	$↗$
$(- 1, \frac{4}{5})$	$0$	$+$	$↗$
$(\frac{4}{5}, 2)$	$1$	$-$	$↘$
$(2, \infty)$	$3$	$+$	$↘$

3. Interpret

From $(\infty, - 1)$ , $g^{'} (x)$ is positive, so $g (x)$ is increasing.

From $(- 1, \frac{4}{5})$ , $g^{'} (x)$ is still positive, so $g (x)$ continues increasing. Since there’s no sign change at $x = - 1$ , there is no relative maximum or minimum at $x = - 1$ .

Around $x = \frac{4}{5}$ , $g^{'} (x)$ changes from positive to negative. So $g (x)$ increases and then decreases, which means there is a relative maximum at $x = \frac{4}{5}$ .

Around $x = 2$ , $g^{'} (x)$ changes from negative to positive. So $g (x)$ decreases and then increases, which means there is a relative minimum at $x = 2$ .

Because $g (x)$ is a polynomial with an odd degree, it has no absolute extrema. As $x \to \infty$ , $g (x) \to \infty$ , and as $x \to - \infty$ , $g (x) \to - \infty$ .

AP tip:

It suffices to make a statement such as “there is a relative maximum at $x = 1$ because $f (x)$ is increasing and then decreasing” or “because $f^{'} (x)$ changes from positive to negative.”

Next, let’s work with a rational function.

Classify the extrema on

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

Solution

(spoiler)

1. Find critical points

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

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

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

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

So the critical points come only from $f^{'} (x) = 0$ , which happens when the numerator is zero. The critical points are $x = 1, - 1$ .

2. Sign chart

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

3. Interpret

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

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

Next, we’ll look at a function where $f^{'} (a)$ is undefined.

Classify the extrema on

$f (x) = ∣ x^{2} - 1∣$

Solution

(spoiler)

Rewriting $f (x)$ as a piecewise function,

$f (x) = ⎩ ⎨ ⎧ x^{2} - 1 - (x^{2} - 1) x^{2} - 1 if x < - 1 if - 1 < x < 1 if x > 1$

This means the derivative is

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 2 x 2 x if x < - 1 if - 1 < x < 1 if x > 1$

$f^{'} (x) = 0$ when $x = 0$ .

For $x = - 1$ :

$Left:$

$x \to - 1^{-} lim f (x) = 2 (- 1) = - 2$

$Right:$

$x \to - 1^{+} lim f (x) = - 2 (- 1) = 2$

The two one-sided values don’t agree, so the derivative doesn’t exist at $x = - 1$ (so $f^{'} (x)$ is undefined there). But $f (- 1)$ exists (it equals 0), so $x = - 1$ is a critical point.

Similarly, for $x = 1$ ,

$Left: x \to 1^{-} lim f (x) = - 2 (1) = - 2$

$Right: x \to 1^{+} lim f (x) = 2 (1) = 2$

Since the two sides don’t agree, $f^{'} (1)$ is undefined and $x = 1$ is also a critical point.

We have three critical points: $x = - 1, 0, 1$ , so there are 4 intervals where we can test the sign of $f^{'} (x)$ .

2. Sign chart

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

3. Interpret

Around $x = - 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Around $x = 0$ , $f^{'} (x)$ changes from positive to negative $\to$ relative maximum.

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Lastly, we’ll discuss critical points that occur periodically. Trig functions often behave this way.

Classify all extrema on

$f (x) = e^{s i n (x)}$

Solution

(spoiler)

1. Find critical points

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

To solve $f^{'} (x) = 0$ , we need

$e^{s i n (x)} = 0$ or
$cos (x) = 0$ .

Since $e^{anything} > 0$ for all real inputs, $e^{s i n (x)}$ is never 0. So the only way for $f^{'} (x) = 0$ is $cos (x) = 0$ .

$cos (x) = 0$ when $x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, etc.$ These critical points repeat every $π$ .

2. Sign chart

Although $f (x)$ repeats forever, we’ll limit the sign chart to these intervals: $(- \frac{π}{2}, \frac{π}{2}), (\frac{π}{2}, \frac{3 π}{2}), (\frac{3 π}{2}, \frac{5 π}{2})$ .

Instead of testing points, we can determine the sign of $f^{'} (x)$ by reasoning:

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

$e^{s i n (x)}$ is always positive.
So the sign of $f^{'} (x)$ is the same as the sign of $cos (x)$ .

On $(- \frac{π}{2}, \frac{π}{2})$ , $cos (x) > 0$ (quadrants I and IV), so $f^{'} (x) > 0$ .

On $(\frac{π}{2}, \frac{3 π}{2})$ , $cos (x) < 0$ (quadrants II and III), so $f^{'} (x) < 0$ .

On $(\frac{3 π}{2}, \frac{5 π}{2})$ , $cos (x) > 0$ again, so $f^{'} (x) > 0$ .

Here is the organized sign chart:

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \frac{π}{2}, \frac{π}{2})$	$+$	$↗$
$(\frac{π}{2}, \frac{3 π}{2})$	$-$	$↘$
$(\frac{3 π}{2}, \frac{5 π}{2})$	$+$	$↗$

3. Interpret

Around $x = \frac{π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

Around $x = \frac{3 π}{2}$ , $f (x)$ decreases and then increases $\to$ relative min.

Around $x = \frac{5 π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

The critical points will continue to alternate between relative maxima and minima.

Extrema

What you’ll learn:

How to find critical points of various functions
How to use the 1st derivative test (and sign charts) to identify relative/local extrema

the endpoints of the interval, and
any critical points inside the interval.

The 1st derivative test gives you a different tool: it helps you classify critical points (anywhere in the domain) as relative/local maxima, relative/local minima, or neither.

Absolute vs. relative extrema

A global (absolute) extremum is the highest or lowest value a function attains over its entire domain.

For example, consider $f (x) = x^{2}$ .

On the other hand, $x^{2}$ has no absolute maximum on its full domain, because as $x \to \pm \infty$ , $x^{2} \to \infty$ .

First derivative test

Here’s how to find and classify extrema using the 1st derivative test:

1. Find critical points by setting $f^{'} (x) = 0$ or by finding where $f^{'} (x)$ is undefined.

Important: $x = a$ is only a critical point if it’s in the domain of $f (x)$ (so $f (a)$ must be defined). The next page will discuss this in more detail, but it’s worth keeping in mind now.

3. Interpret each critical point

If $f^{'} (x)$ changes from positive to negative at a critical point, then $f (x)$ increases ( $↗$ ) and then decreases ( $↘$ ). The critical point is a local/relative maximum (a peak).
If $f^{'} (x)$ changes from negative to positive at a critical point, then $f (x)$ decreases ( $↘$ ) and then increases ( $↗$ ). The critical point is a local/relative minimum (a valley).
If there is no sign change, the critical point is not an extremum. For example, $y = x^{3}$ is flat at $(0, 0)$ , but the function keeps increasing through that point.

Examples

We’ll start with a basic polynomial.

Classify the extrema on

$g (x) = (x - 2)^{2} (x + 1)^{3}$

Solution

1. Find critical points

Using the product rule,

$g^{'} (x) = (x - 2)^{2} \cdot 3 (x + 1)^{2} + (x + 1)^{3} \cdot 2 (x - 2)$

$= (x - 2) (x + 1)^{2} [3 (x - 2) + 2 (x + 1)]$

$= (x - 2) (x + 1)^{2} (5 x - 4)$

Since $g^{'} (x)$ is a polynomial, it’s defined for all real $x$ . To find critical points, solve $g^{'} (x) = 0$ :

$(x - 2) (x + 1)^{2} (5 x - 4) = 0$

$x = 2, - 1, \frac{4}{5}$

2. Create a sign diagram or chart

Here it’ll be displayed as a sign chart, but a sign diagram with the intervals displayed left to right is usually easier to read (especially when you graph later).

Interval	Test point $x =$	Sign of $g^{'} (x)$	$g (x)$ behavior
$(\infty, - 1)$	$- 2$	$+$	$↗$
$(- 1, \frac{4}{5})$	$0$	$+$	$↗$
$(\frac{4}{5}, 2)$	$1$	$-$	$↘$
$(2, \infty)$	$3$	$+$	$↘$

3. Interpret

From $(\infty, - 1)$ , $g^{'} (x)$ is positive, so $g (x)$ is increasing.

From $(- 1, \frac{4}{5})$ , $g^{'} (x)$ is still positive, so $g (x)$ continues increasing. Since there’s no sign change at $x = - 1$ , there is no relative maximum or minimum at $x = - 1$ .

Around $x = \frac{4}{5}$ , $g^{'} (x)$ changes from positive to negative. So $g (x)$ increases and then decreases, which means there is a relative maximum at $x = \frac{4}{5}$ .

Around $x = 2$ , $g^{'} (x)$ changes from negative to positive. So $g (x)$ decreases and then increases, which means there is a relative minimum at $x = 2$ .

Because $g (x)$ is a polynomial with an odd degree, it has no absolute extrema. As $x \to \infty$ , $g (x) \to \infty$ , and as $x \to - \infty$ , $g (x) \to - \infty$ .

AP tip:

It suffices to make a statement such as “there is a relative maximum at $x = 1$ because $f (x)$ is increasing and then decreasing” or “because $f^{'} (x)$ changes from positive to negative.”

Next, let’s work with a rational function.

Classify the extrema on

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

Solution

(spoiler)

1. Find critical points

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

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

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

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

So the critical points come only from $f^{'} (x) = 0$ , which happens when the numerator is zero. The critical points are $x = 1, - 1$ .

2. Sign chart

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

3. Interpret

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

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

Next, we’ll look at a function where $f^{'} (a)$ is undefined.

Classify the extrema on

$f (x) = ∣ x^{2} - 1∣$

Solution

(spoiler)

Rewriting $f (x)$ as a piecewise function,

$f (x) = ⎩ ⎨ ⎧ x^{2} - 1 - (x^{2} - 1) x^{2} - 1 if x < - 1 if - 1 < x < 1 if x > 1$

This means the derivative is

$f^{'} (x) = ⎩ ⎨ ⎧ 2 x - 2 x 2 x if x < - 1 if - 1 < x < 1 if x > 1$

$f^{'} (x) = 0$ when $x = 0$ .

For $x = - 1$ :

$Left:$

$x \to - 1^{-} lim f (x) = 2 (- 1) = - 2$

$Right:$

$x \to - 1^{+} lim f (x) = - 2 (- 1) = 2$

The two one-sided values don’t agree, so the derivative doesn’t exist at $x = - 1$ (so $f^{'} (x)$ is undefined there). But $f (- 1)$ exists (it equals 0), so $x = - 1$ is a critical point.

Similarly, for $x = 1$ ,

$Left: x \to 1^{-} lim f (x) = - 2 (1) = - 2$

$Right: x \to 1^{+} lim f (x) = 2 (1) = 2$

Since the two sides don’t agree, $f^{'} (1)$ is undefined and $x = 1$ is also a critical point.

We have three critical points: $x = - 1, 0, 1$ , so there are 4 intervals where we can test the sign of $f^{'} (x)$ .

2. Sign chart

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

3. Interpret

Around $x = - 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Around $x = 0$ , $f^{'} (x)$ changes from positive to negative $\to$ relative maximum.

Around $x = 1$ , $f^{'} (x)$ changes from negative to positive $\to$ relative minimum.

Lastly, we’ll discuss critical points that occur periodically. Trig functions often behave this way.

Classify all extrema on

$f (x) = e^{s i n (x)}$

Solution

(spoiler)

1. Find critical points

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

To solve $f^{'} (x) = 0$ , we need

$e^{s i n (x)} = 0$ or
$cos (x) = 0$ .

Since $e^{anything} > 0$ for all real inputs, $e^{s i n (x)}$ is never 0. So the only way for $f^{'} (x) = 0$ is $cos (x) = 0$ .

$cos (x) = 0$ when $x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, etc.$ These critical points repeat every $π$ .

2. Sign chart

Although $f (x)$ repeats forever, we’ll limit the sign chart to these intervals: $(- \frac{π}{2}, \frac{π}{2}), (\frac{π}{2}, \frac{3 π}{2}), (\frac{3 π}{2}, \frac{5 π}{2})$ .

Instead of testing points, we can determine the sign of $f^{'} (x)$ by reasoning:

$f^{'} (x) = e^{s i n (x)} \cdot cos (x)$

$e^{s i n (x)}$ is always positive.
So the sign of $f^{'} (x)$ is the same as the sign of $cos (x)$ .

On $(- \frac{π}{2}, \frac{π}{2})$ , $cos (x) > 0$ (quadrants I and IV), so $f^{'} (x) > 0$ .

On $(\frac{π}{2}, \frac{3 π}{2})$ , $cos (x) < 0$ (quadrants II and III), so $f^{'} (x) < 0$ .

On $(\frac{3 π}{2}, \frac{5 π}{2})$ , $cos (x) > 0$ again, so $f^{'} (x) > 0$ .

Here is the organized sign chart:

Interval	Sign of $f^{'} (x)$	$f (x)$ behavior
$(- \frac{π}{2}, \frac{π}{2})$	$+$	$↗$
$(\frac{π}{2}, \frac{3 π}{2})$	$-$	$↘$
$(\frac{3 π}{2}, \frac{5 π}{2})$	$+$	$↗$

3. Interpret

Around $x = \frac{π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

Around $x = \frac{3 π}{2}$ , $f (x)$ decreases and then increases $\to$ relative min.

Around $x = \frac{5 π}{2}$ , $f (x)$ increases and then decreases $\to$ relative max.

The critical points will continue to alternate between relative maxima and minima.