Extrema | 1st derivative test | Analytical uses

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, we introduced the notion of critical points and finding the highest and lowest values on a closed interval using the extreme value theorem. It required comparing function values at the endpoints and critical points to determine the absolute maximum and minimum.

On the other hand, using the 1st derivative test allows you to find and classify any critical point over a function’s domain as a relative maximum or a minimum.

Absolute vs. relative extrema

A global extrema is the absolute maximum or minimum value of a function.

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

The vertex $(0, 0)$ is the global/absolute minimum because it’s the lowest point across the domain of all real numbers. But it’s also a relative/local minimum because relative to its immediate surrounding points, it’s lower. On the other hand, $x^{2}$ has no absolute maximum on the entire domain because it increases without bound to $\infty$ on both sides.

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 determining where it is undefined.

Important: $x = a$ is only a critical point if it’s in the domain of $f (x)$ (so if $f (a)$ is defined). The next page will discuss this in more detail but for now we’ll just mention it in advance.

2. Create a sign diagram or chart for $f^{'} (x)$ with regions/intervals around each critical point. Look at 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 around the critical point, it means that $f (x)$ increases ( $↗$ ) and then decreases ( $↘$ ), so the critical point is a local/relative maximum (picture a peak).
If $f^{'} (x)$ changes from negative to positive, it means that $f (x)$ decreases ( $↘$ ) and then increases ( $↗$ ), so the critical point is a local/relative minimum (picture a valley).
If there is no sign change, the critical point is not an extremum. For example, the graph of $y = x^{3}$ is flat at $(0, 0)$ but the graph continues moving up instead of appearing as a peak or a valley.

Examples

We’ll start with a basic polynomial.

1. 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 also a polynomial, there is no point where it’s undefined. Solving for $g^{'} (x) = 0$ ,

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

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

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 look at, especially for graphing later on.

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 $(- 1, \frac{4}{5})$ , $g (x)$ continues increasing. This means that 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 $x$ becomes very large, $y$ also follows. As $x$ approaches $- \infty$ , so does $y$ . :::

AP tip:

In addition to a sign chart for $f^{'} (x)$ , you must state what it means 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.

2. Classify the extrema on

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

Solution

(spoiler)

1. Find critical points

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

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

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

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

Usually, rational functions are undefined when the denominator is zero. But the denominator of $f^{'} (x)$ is never undefined because $(x^{2} + 1)^{2} > 0$ for all $x$ . So the only critical points are when $f^{'} (x) = 0$ , or when the numerator is $0$ . 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 the graph of $f (x)$ decreases and then increases and there is a relative minimum at $x = - 1$ .

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

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

3. 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 (or where it’s not differentiable), we have to revisit the concepts from section 2.6 on differentiability and continuity.

When $f (x)$ is a piecewise function, confirm that the derivatives approaching from either side of a breakpoint are the same value (otherwise the graph looks choppy rather than smooth, and there is no single value for the derivative).

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 sides don’t agree, so the derivative doesn’t exist at $x = - 1$ ( $f^{'} (x)$ 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: $x = - 1, 0, 1$ and 4 intervals to test points in.

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. Usually trig functions are involved.

4. Classify all extrema on

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

Solution

(spoiler)

1. Find critical points

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

Solving $f^{'} (x) = 0$ , either $e^{s i n (x)} = 0$ or $cos (x) = 0$ .

Since $e^{anything} > 0$ for all $x$ and the domain of $sin (x)$ is all real numbers, it’s defined for all $x$ and never equals $0$ .

$cos (x)$ is also defined for all real numbers but it equals $0$ when $x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, etc.$

The critical points repeat in increments of $π$ to the left and right of $\frac{π}{2}$ .

2. Sign chart

Although $f (x)$ is periodic and repeats forever, we will 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, let’s reason through the sign of $f^{'} (x)$ in each interval again. You may, of course, test the suggested points to confirm.

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

$e^{s i n (x)}$ is always positive. The interval $(- \frac{π}{2}, \frac{π}{2})$ corresponds to the angle $x$ in quadrants I & IV (upper and lower right) where $cos (x)$ is positive. You may test a point like $x = 0$ .

The next interval $(\frac{π}{2}, \frac{3 π}{2})$ corresponds to quadrants II & III (upper and lower left) where $cos (x)$ is negative. You may test a point like $x = π$ .

The interval $(\frac{3 π}{2}, \frac{5 π}{2})$ corresponds again to quadrants I & IV where $cos (x)$ is positive. You may test a point like $x = 2 π$ .

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 alternate between relative maximums and minimums onwards.

Key points

Find critical points by setting $f^{'} (x) = 0$ or finding where $f^{'} (x)$ is undefined.
Use a sign diagram or chart to display the sign of $f^{'} (x)$ in each interval around the critical points.
Interpret the behavior of $f (x)$ (increasing or decreasing) based on the sign of $f^{'} (x)$ to justify if a critical point is a relative max or min.

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

On the other hand, using the 1st derivative test allows you to find and classify any critical point over a function’s domain as a relative maximum or a minimum.

Absolute vs. relative extrema

A global extrema is the absolute maximum or minimum value of a function.

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

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 determining where it is undefined.

Important: $x = a$ is only a critical point if it’s in the domain of $f (x)$ (so if $f (a)$ is defined). The next page will discuss this in more detail but for now we’ll just mention it in advance.

3. Interpret each critical point

If $f^{'} (x)$ changes from positive to negative around the critical point, it means that $f (x)$ increases ( $↗$ ) and then decreases ( $↘$ ), so the critical point is a local/relative maximum (picture a peak).
If $f^{'} (x)$ changes from negative to positive, it means that $f (x)$ decreases ( $↘$ ) and then increases ( $↗$ ), so the critical point is a local/relative minimum (picture a valley).
If there is no sign change, the critical point is not an extremum. For example, the graph of $y = x^{3}$ is flat at $(0, 0)$ but the graph continues moving up instead of appearing as a peak or a valley.

Examples

We’ll start with a basic polynomial.

1. 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 also a polynomial, there is no point where it’s undefined. Solving for $g^{'} (x) = 0$ ,

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

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

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 look at, especially for graphing later on.

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 $(- 1, \frac{4}{5})$ , $g (x)$ continues increasing. This means that 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 $x$ becomes very large, $y$ also follows. As $x$ approaches $- \infty$ , so does $y$ . :::

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.

2. Classify the extrema on

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

Solution

(spoiler)

1. Find critical points

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

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

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

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

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 the graph of $f (x)$ decreases and then increases and there is a relative minimum at $x = - 1$ .

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

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

3. 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 (or where it’s not differentiable), we have to revisit the concepts from section 2.6 on differentiability and continuity.

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 sides don’t agree, so the derivative doesn’t exist at $x = - 1$ ( $f^{'} (x)$ 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: $x = - 1, 0, 1$ and 4 intervals to test points in.

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. Usually trig functions are involved.

4. Classify all extrema on

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

Solution

(spoiler)

1. Find critical points

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

Solving $f^{'} (x) = 0$ , either $e^{s i n (x)} = 0$ or $cos (x) = 0$ .

Since $e^{anything} > 0$ for all $x$ and the domain of $sin (x)$ is all real numbers, it’s defined for all $x$ and never equals $0$ .

$cos (x)$ is also defined for all real numbers but it equals $0$ when $x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, etc.$

The critical points repeat in increments of $π$ to the left and right of $\frac{π}{2}$ .

2. Sign chart

Although $f (x)$ is periodic and repeats forever, we will 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, let’s reason through the sign of $f^{'} (x)$ in each interval again. You may, of course, test the suggested points to confirm.

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

The next interval $(\frac{π}{2}, \frac{3 π}{2})$ corresponds to quadrants II & III (upper and lower left) where $cos (x)$ is negative. You may test a point like $x = π$ .

The interval $(\frac{3 π}{2}, \frac{5 π}{2})$ corresponds again to quadrants I & IV where $cos (x)$ is positive. You may test a point like $x = 2 π$ .

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 alternate between relative maximums and minimums onwards.

Key points

Find critical points by setting $f^{'} (x) = 0$ or finding where $f^{'} (x)$ is undefined.
Use a sign diagram or chart to display the sign of $f^{'} (x)$ in each interval around the critical points.
Interpret the behavior of $f (x)$ (increasing or decreasing) based on the sign of $f^{'} (x)$ to justify if a critical point is a relative max or min.