Question

Use (a) the $h\to 0$definition of the derivative and then

(b) the $z\to c$definition of the derivative to find $f\text{'}\left(c\right)$for each function f and value $x=c$ in Exercises 23–38.

23.$f\left(x\right)=x^2,x=-3$

Short answer

$f\text{'}(-3)=-6$.

Step-by-step solution

Part (a) Step 1. Given information.

A function is given as$f\left(x\right)=x^2$and$x=c=-3$.

Part (a) Step 2. Find f'(c) using h→0 definition of the derivative.

We have

$$\begin{gathered} f\text{'}(-3)=\underset{h\to 0}{\lim }\frac{f(-3+h)-f(-3)}{h} \\ =\underset{h\to 0}{\lim }\frac{{(-3+h)}^2-{(-3)}^2}{h} \\ =\underset{h\to 0}{\lim }\frac{9-6h+h^2-9}{h} \\ =\underset{h\to 0}{\lim }\frac{-6h+h^2}{h} \\ =\underset{h\to 0}{\lim }\frac{h(h-6)}{h} \\ =\underset{h\to 0}{\lim }(h-6) \\ =0-6 \\ =-6 \end{gathered}$$

Part (b) Step 2. Find f'(c) using z→(-3) definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c} \\ f\text{'}(-3)=\underset{z\to -3}{\lim }\frac{z^2-{(-3)}^2}{z-(-3)} \\ =\underset{z\to -3}{\lim }\frac{z^2-9}{z+3} \\ =\underset{z\to -3}{\lim }\frac{(z-3)(z+3)}{(z+3)} \\ =\underset{z\to -3}{\lim }(z-3) \\ =\left[\right(-3)-3] \\ =-6 \end{gathered}$$