Find critical numbers

A critical number is an input $c$ in the domain of a differentiable-function model for which $f'(c)=0$ or $f'(c)$ does not exist, such as at a corner, cusp, or vertical tangent; points where the function is undefined are not critical numbers. Critical numbers identify candidates for local maxima and minima and are analyzed with derivative sign changes, while endpoints are considered separately for absolute extrema. This scope is limited to single-variable functions and does not include advanced generalized definitions.

Detailed Explanation: Find critical numbers

A critical number is an input $c$ in the domain of $f$ where

$(f'(c)=0)$ , or
$(f'(c))$ does not exist.

A point where $f$ is undefined is not a critical number.

Worked example

Find the critical numbers of

f(x)= \vert x-2 \vert +(x-3)^2.

Step 1: Find the domain

Both $( |x-2| )$ and $( (x-3)^2 )$ are defined for every real number. Therefore,

\text{Domain of }f=(-\infty,\infty).

Step 2: Rewrite the absolute value using cases

The expression $( |x-2| )$ changes form at $(x=2)$ :

\vert x-2 \vert = \begin{cases} -(x-2), & x<2,\\ x-2, & x>2. \end{cases}

So we differentiate on each side of $(x=2)$ :

For $(x<2)$ ,

f(x)=-(x-2)+(x-3)^2,

f'(x)=-1+2(x-3)=2x-7.

For $(x>2)$ ,

f(x)=(x-2)+(x-3)^2,

f'(x)=1+2(x-3)=2x-5.

Step 3: Find where $(f'(x)=0)$

On the interval $(x<2)$ ,

2x-7=0 \quad\Rightarrow\quad x=\frac{7}{2}.

But $\frac{7}{2}>2$ , so it is not in the interval $(x<2)$ . It gives no critical number from this part.

On the interval $(x>2)$ ,

2x-5=0 \quad\Rightarrow\quad x=\frac{5}{2}.

Since $\frac{5}{2}>2$ , this value is valid. Thus,

x=\frac{5}{2}

is a critical number.

Step 4: Check where $(f'(x))$ does not exist

The formula changes at $(x=2)$ , so check the one-sided derivatives:

f'_-(2)=2(2)-7=-3,

and

f'_+(2)=2(2)-5=-1.

Because the one-sided derivatives are different, $(f'(2))$ does not exist. Since $(f(2))$ is defined, $(x=2)$ is also a critical number.

Therefore, the critical numbers are

\boxed{x=2 \text{ and } x=\frac{5}{2}}.

These values are candidates for local maxima or minima. They must be analyzed further, usually by checking whether the derivative changes sign.

Learn by doing: Find critical numbers

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Critical Points - Function to Critical Numbers

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