Chapter 4.2, Problem 23E

To determine

To find: The critical numbers of the function $f\left(x\right)=4+\frac{1}{3}x-\frac{1}{2}{x}^2$.

Answer to Problem 23E

The critical number of $f\left(x\right)=4+\frac{1}{3}x-\frac{1}{2}{x}^2$ is, $x=\frac{1}{3}$.

Explanation of Solution

Definition used:

A critical number of a function f is a number c, if it satisfies either of the below conditions:

(1) $f^{\prime }\left(c\right)=0$

(2) $f^{\prime }\left(c\right)$ does not exist.

Formula used:

The Power Rule:

If n is positive integer, then $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$. (1)

Calculation:

Obtain the first derivative of the given function $f\left(x\right)=4+\frac{1}{3}x-\frac{1}{2}{x}^2$.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(4+\frac{1}{3}x-\frac{1}{2}{x}^2\right)\\ =\frac{d}{dx}\left(4\right)+\frac{d}{dx}\left(\frac{1}{3}x\right)-\frac{d}{dx}\left(\frac{1}{2}{x}^2\right)\\ =0+\frac{1}{3}\left(1\right)-\frac{1}{2}\left(2\right)x^{2-1}\left[\text{Apply power rule as in the equation }\left(1\right)\right]\\ =\frac{1}{3}-x\end{array}$

Take $f^{\prime }\left(x\right)=0$ to obtain the critical point.

$\begin{array}{l}\frac{1}{3}-x=0\\ x=\frac{1}{3}\end{array}$

Thus, the critical number of $f\left(x\right)=4+\frac{1}{3}x-\frac{1}{2}{x}^2$ is, $x=\frac{1}{3}$.