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# Critical numbers – Definition, Process, and Examples

Knowing how to find a function’s **critical numbers **will come in handy when we want to determine extreme values and quantities. Finding critical numbers is a helpful optimization technique applied in physics, finance, and engineering as well.

*Critical numbers are values of a function where the function’s tangent lines are either horizontal or vertical. *

In this article, we’ll focus on critical numbers that return values of $x$ where $f’(x)$ is either zero or an undefined value. We’ll also learn how to identify them and understand what these points represent.

**What are critical numbers?**

Critical numbers or critical points are values of $x$ where the **first derivative of a function is either equal to zero or undefined**. Let’s say we have $x = c$, the critical numbers of the function,$f(x)$, will satisfy either of the following:

\begin{aligned}f’(c) &= 0 \\\\f’(c) &= \text{ DNE (Does Not Exist)}\end{aligned}

This means that $x =c$ is a critical number when a **tangent line passing through** $\boldsymbol{x = c}$ **is either a horizontal or vertical line.**

The graph above shows us examples of critical numbers meeting different conditions. Let’s break down what each critical number represents:

- The
**local extremums**(both minimum and maximum) indicate the**extremum value within an interval**. - The
**global extremum**tells us the definite maximum or minimum value of the function**throughout its domain**. - Points where the function
**“plateau” or inflect**tells us that the derivative changes from**increasing to decreasing (or the other way around)**.

All these components will only be considered critical points of the function when the **first derivative is either zero or doesn’t exist**.

**How to find critical numbers? **

We can use the definition of critical numbers to determine the critical numbers – by finding the values of $x$ where $f’(x) = 0$ or $f’(x) = \text{ DNE}$. These are some guide points to help you find a function’s critical numbers:

- Use the derivative rules to find $f’(x)$’s expression.
- Find the values of $x$ where $f’(x) = 0$.
- Include the values of $x$ where $f’(x)$ is undefined.

Only include critical numbers that are within the function’s domain. This means that values of $x$ where $f’(x)$ is undefined will only count when the point is within the domain of $f(x)$.

Now, take a look at this graph and observe how these critical numbers would appear on an $xy$-coordinate system.

When given the graph of $f(x)$, we can inspect the points where the tangent line passing through $f(x)$ is either vertical or horizontal.

- When the tangent line at $x = c$ is vertical, $f’(c) = 0$ and we have critical point at $x = c$.
- Similarly, when the tangent line passing through $x = c$ is horizontal, $f’(c)$ is undefined and $x=c$ is also a critical point.

Let’s use these techniques and tips to find the critical numbers of the functions shown below.

*Example 1*

What are the critical numbers of the function, $f(x) = 2x^3 – 8x^2 + 2x – 1$?

__Solution__

We can determine the critical numbers of $f(x)$ by first finding the expression for $f(x)$’s derivative.

- Use the sum and difference rules to distribute $\dfrac{d}{dx}$ throughout each term.
- Factor out the coefficient by applying the constant multiple rule.
- Use the power rule to simplify the terms of $f’(x)$.

\begin{aligned}f(x) &= 2x^3 – 8x^2 + 2x – 1\\f'(x) &= \dfrac{d}{dx}2x^3 – \dfrac{d}{dx}8x^2 + \dfrac{d}{dx}2x – \dfrac{d}{dx}1,\phantom{x}{\color{Orchid} \text{Sum and Difference Rules}}\\&= {\color{Orchid}2\dfrac{d}{dx}x^3} – {\color{Orchid}8\dfrac{d}{dx}x^2} + {\color{Orchid}2\dfrac{d}{dx}x }- {\color{DarkOrange} 0},\phantom{x}{\color{Orchid} \text{Constant Multiple Rule}}\text{ & }{\color{DarkOrange} \text{Constant Rule}}\\&= 2({\color{Orchid}3x^2 })-8({\color{Orchid}2x })+2({\color{Orchid}1 }),\phantom{x}{\color{Orchid} \text{Power Rule}}\\&= 6x^2 – 16x + 2\end{aligned}

Recall that critical numbers occur at values of $x$ when $f’(x) = 0$ or when $f’(x)$ does not exist. Since $f’(x)$ is a quadratic expression, $x$ has no restrictions. Equate $f’(x)$ to zero to find the critical numbers instead.

\begin{aligned}f'(x) &= 0\\ 6x^2 – 16x + 2 &=0 \\3x^2 – 8x + 1 &= 0\end{aligned}

Since we can’t factor the quadratic expression, we can use the quadratic formula to determine the critical numbers of $f(x)$ instead.

\begin{aligned}x&= \dfrac{-(-8) \pm \sqrt{(-8)^2 – 4(3)(1)}}{2(3)}\\&= \dfrac{8 \pm \sqrt{64 – 12}}{6}\\&= \dfrac{8 \pm \sqrt{52}}{6}\\&=\dfrac{8 \pm 2\sqrt{13}}{6}\\&=\dfrac{4 \pm \sqrt{13}}{3}\end{aligned}

This means that $f(x)$ has critical numbers at $x = \dfrac{4 \pm \sqrt{13}}{3}$.

The graph of $f(x) = 2x^3 – 8x^2 + 2x – 1$ is shown below for you to understand what these critical numbers represent.

We can see that both critical numbers are considered $f(x)$’s local minima and the tangent lines passing through them will be horizontal.

*Example 2*

What are the critical numbers of the function, $g(x) = \dfrac{x^2 – 6}{x -4}$?

__Solution__

We begin by differentiating $g(x)$ and to do so, we’ll apply the following derivative rules.

- Since $g(x)$ is a rational expression, we’ll apply the quotient rule
- Apply the constant and power rules whenever necessary.

\begin{aligned}g(x) &= \dfrac{x^2 – 6}{x -4}\\g'(x) &= \dfrac{(x- 4)\dfrac{d}{dx} (x^2 -6) – (x^2 – 6)\dfrac{d}{dx}(x – 4)}{(x – 4)^2},\phantom{x}{\color{Orchid}\text{Quotient Rule}}\\&= \dfrac{(x- 4){\color{Orchid}\left(\dfrac{d}{dx}x^2 -\dfrac{d}{dx}6\right)} – (x^2 -6){\color{Orchid}\left(\dfrac{d}{dx}x – \dfrac{d}{dx}4\right)}}{(x – 4)^2},\phantom{x}{\color{Orchid}\text{Difference Rule}}\\&=\dfrac{(x- 4)({\color{Orchid}2x} – {\color{DarkOrange}0})- (x^2 – 6)({\color{Orchid}1} – {\color{DarkOrange}0})}{(x – 4)^2},\phantom{x}{\color{Orchid}\text{Power Rule}}\text{ & }{\color{DarkOrange}\text{Constant Rule}}\end{aligned}

Expand the terms in the numerator and combine the terms whenever possible to simplify $g’(x)$.

\begin{aligned}g'(x) &= \dfrac{2x(x -4) – (x^2 – 6)}{(x -4)^2}\\&= \dfrac{2x^2 – 8x – x^2 +6}{(x -4)^2}\\&= \dfrac{x^2 – 8x +6}{(x -4)^2}\end{aligned}

We can find the critical numbers of $g(x)$ by equating $g’(x)$ to zero or by finding undefined values within its domain. For $g’(x)$ to be undefined, the denominator must be zero.

\begin{aligned}(x – 4)^2 &= 0 \\x &= 4\end{aligned}

It may be tempting to include $x = 4$ as one of $g(x)$’s critical numbers, but keep in mind $x = 4$ is not part of $g(x)$’s domain. Hence, we disregard it.

Let’s now focus on the values of $x$ that returns $g’(x) = 0$. After equating the numerator to $0$, apply the quadratic formula to find $x$.

\begin{aligned}g'(x) &=0\\ \dfrac{x^2 – 8x +6}{(x -4)^2}&= 0\\x^2 – 8x + 6 &= 0\\x&= \dfrac{-(-8) \pm \sqrt{(-8)^2 – 4(1)(6)}}{2(1)}\\&=\dfrac{8 \pm \sqrt{64 -24}}{2}\\&=\dfrac{8 \pm 2\sqrt{10}}{2}\\&= 4 \pm \sqrt{10}\end{aligned}

This means that $g(x)$ has critical numbers at $x = 4 – \sqrt{10}$ and $x = 4 + \sqrt{10}$.

Here’s the graph of $g(x)$ to show you where the critical numbers are located. From this, we can see that $x = 4 – \sqrt{10}$ is a local maximum while $x = 4+\sqrt{10}$ is a local minimum.

**Practice Questions**

1. What are the critical numbers of the following functions?

a. $f(x) = 2x^4 – 5x^3 – 6x^2$

b. $g(x) = -5x^3 – 4x^2 + 12$

c. $h(x) = 12x^6 – 8x^4 + 202$

2. What are the critical numbers of the following functions?

a. $f(x) = \dfrac{x – 1}{x^2 + 4}$

b. $g(x) = \dfrac{x^2 – 3}{x – 2}$

c. $h(x) = (x +3)^2 (x -2)$

**Answer Key**

1.

a. $x = \left\{\dfrac{-15 -\sqrt{609}}{3}, 0 ,\dfrac{15 – \sqrt{609}}{3} \right\}$

b. $x=\left\{-\dfrac{18}{5},0\right\}$

c. $x=\left\{-\dfrac{2\sqrt{3}}{3}, 0, \dfrac{2\sqrt{3}}{3}\right\}$

2.

a. $x = \{1 – \sqrt{5}, 1 + \sqrt{5}\}$

b. $x = \{1, 3\}$

c. $x=\left\{-3, \dfrac{1}{3}\right\}$

*Images/mathematical drawings are created with GeoGebra.*