On critical points | 1st derivative test | Analytical uses

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. Here’s the key detail to keep in mind: $f^{'} (a)$ only matters if $a$ is in the domain of $f$ .

So even if $f^{'} (a) = 0$ or $f^{'} (a)$ is undefined, $x = a$ is not a critical point unless $f (a)$ exists. Let’s look at examples where “possible” critical points get eliminated because the original function isn’t defined there.

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 $x$ , 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 test intervals to the right of $0$ .

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 $x = 1$ .

Here’s another example where a potential critical point is eliminated because the function is discontinuous.

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

$f^{'} (x)$ is undefined when $x = 1$ , but $f (x)$ is also undefined at $x = 1$ . Since $x = 1$ is not in the domain of $f$ , 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. 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

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 “possible” 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 $f (x)$ can differ on either side.

Classify the extrema on

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

Solution

(spoiler)

1. Find critical points

The derivative is

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

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

$f^{'} (x) = 0$ 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 $x = 1$ .

$x = a$ is a critical point if $f^{'} (a) = 0$ or is undefined but only if $f (a)$ is defined.

What you’ll learn:

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

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 $x$ , 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 test intervals to the right of $0$ .

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 $x = 1$ .

Here’s another example where a potential critical point is eliminated because the function is discontinuous.

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

$f^{'} (x)$ is undefined when $x = 1$ , but $f (x)$ is also undefined at $x = 1$ . Since $x = 1$ is not in the domain of $f$ , 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. 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

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

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

$f^{'} (x) = 0$ 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 $x = 1$ .

$x = a$ is a critical point if $f^{'} (a) = 0$ or is undefined but only if $f (a)$ is defined.

What you’ll learn:

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

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 $x$ , 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 test intervals to the right of $0$ .

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 $x = 1$ .

Here’s another example where a potential critical point is eliminated because the function is discontinuous.

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

$f^{'} (x)$ is undefined when $x = 1$ , but $f (x)$ is also undefined at $x = 1$ . Since $x = 1$ is not in the domain of $f$ , 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. 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

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

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

$f^{'} (x) = 0$ 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 $x = 1$ .

$x = a$ is a critical point if $f^{'} (a) = 0$ or is undefined but only if $f (a)$ is defined.