On critical points | 1st derivative test | Analytical uses

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 for various functions by setting $f^{'} (x) = 0$ or determining where it’s undefined. But it’s important to remember that the derivative only exists for values of $x$ that are in the domain of the original function. Let’s take a look at a situation where $f^{'} (a) = 0$ or undefined but $f (a)$ does not exist, which means that $x = a$ isn’t actually a critical point and can’t be a relative extrema.

1. 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 be the same as the rational function in example 2 on the previous page. The derivative, with the chain rule, 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 both top and bottom and simplifying,

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

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

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

However, the original function only exists where the argument $\frac{x}{x ^{2} + 1}$ is greater than $0$ , or when $x > 0$ . So the derivative is also only valid in the domain of the original function.

If a point is not defined on the original function, then it can’t be a critical point. Neither $f (0)$ nor $f (- 1)$ are defined.

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

2. Sign chart

Because the domain of $f (x)$ is $x > 0$ , we don’t need to test points less than $0$ .

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. So $f (x)$ increases and then decreases and there is a relative maximum at $x = 1$ .

Here’s another example where a potential critical point is eliminated due to a discontinuity.

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

$f^{'} (x)$ is undefined when $x = 1$ , but $x = 1$ was already undefined in the original function $f (x)$ , so it isn’t a critical point. Therefore this function has no critical points. Which is perfectly fine - not every function will have critical points!

However, we can still analyze the graph using a sign chart, a technique that will be important in section 5.4 on curve sketching.

2. Sign chart

In this part, we’ll skip testing points in order to show how to reason through the behavior of $f (x)$ just by considering $f^{'} (x)$ . Because the denominator of $f^{'} (x)$ is squared, it’s always positive. But the negative sign in front of the fraction means that $f^{'} (x) < 0$ for all $x$ in its domain. So the sign chart is:

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

3. Interpret

There are no critical points so no relative maximum/minimums. The asymptote at $x = 1$ means that the function approaches $\infty$ and $- \infty$ on either side, so there is no absolute maximum or minimum either.

Even if a potential critical point is a discontinuity, it’s still important to include it in the sign chart since it splits the domain of the function. It won’t be a relative extrema but the behavior of $f (x)$ may be different on either side. Here is an example of this situation:

3. 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 and will never equal $0$ . Solving $1 - \frac{1}{x} = 0$ ,

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

$x = 1$

$f^{'} (x)$ is undefined when $x = 0$ . But since $f (x)$ is also undefined there, $x = 0$ is not in the domain of the original function and isn’t a critical point.

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

2. Sign chart

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

3. Interpret

$f (x)$ is not continuous at $x = 0$ so even though $f^{'} (x)$ changes from positive to negative, it’s not a relative max nor min.

Around $x = 1$ , So $f (x)$ decreases and then increases $\to$ relative min.

Key points

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

On critical points

What you’ll learn:

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

1. 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 be the same as the rational function in example 2 on the previous page. The derivative, with the chain rule, 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 both top and bottom and simplifying,

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

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

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

If a point is not defined on the original function, then it can’t be a critical point. Neither $f (0)$ nor $f (- 1)$ are defined.

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

2. Sign chart

Because the domain of $f (x)$ is $x > 0$ , we don’t need to test points less than $0$ .

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. So $f (x)$ increases and then decreases and there is a relative maximum at $x = 1$ .

Here’s another example where a potential critical point is eliminated due to a discontinuity.

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

However, we can still analyze the graph using a sign chart, a technique that will be important in section 5.4 on curve sketching.

2. Sign chart

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

3. Interpret

3. 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 and will never equal $0$ . Solving $1 - \frac{1}{x} = 0$ ,

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

$x = 1$

$f^{'} (x)$ is undefined when $x = 0$ . But since $f (x)$ is also undefined there, $x = 0$ is not in the domain of the original function and isn’t a critical point.

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

2. Sign chart

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

3. Interpret

$f (x)$ is not continuous at $x = 0$ so even though $f^{'} (x)$ changes from positive to negative, it’s not a relative max nor min.

Around $x = 1$ , So $f (x)$ decreases and then increases $\to$ relative min.

Key points

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