Question

use the four - step definition of the derivative to find $f(x)$ if $f(x)=sqrt{-7x - 18}$. you may have to simplify your answers to recieve credit.
$f(x + h)=
$f(x + h)-f(x)=
$\frac{f(x + h)-f(x)}{h}=
$f(x)=$

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$:
$f(x + h)=\sqrt{-7(x + h)-18}=\sqrt{-7x-7h - 18}$

Step2: Calculate $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=\sqrt{-7x-7h - 18}-\sqrt{-7x - 18}\\ \end{align*}$$

\]

Step3: Compute $\frac{f(x + h)-f(x)}{h}$

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{\sqrt{-7x-7h - 18}-\sqrt{-7x - 18}}{h}\\ &=\frac{(\sqrt{-7x-7h - 18}-\sqrt{-7x - 18})(\sqrt{-7x-7h - 18}+\sqrt{-7x - 18})}{h(\sqrt{-7x-7h - 18}+\sqrt{-7x - 18})}\\ &=\frac{(-7x-7h - 18)-(-7x - 18)}{h(\sqrt{-7x-7h - 18}+\sqrt{-7x - 18})}\\ &=\frac{-7x-7h - 18 + 7x+18}{h(\sqrt{-7x-7h - 18}+\sqrt{-7x - 18})}\\ &=\frac{-7h}{h(\sqrt{-7x-7h - 18}+\sqrt{-7x - 18})}\\ &=\frac{-7}{\sqrt{-7x-7h - 18}+\sqrt{-7x - 18}} \end{align*}$$

\]

Step4: Find $f'(x)$

Take the limit as $h
ightarrow0$:
\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}\frac{-7}{\sqrt{-7x-7h - 18}+\sqrt{-7x - 18}}\\ &=\frac{-7}{\sqrt{-7x - 18}+\sqrt{-7x - 18}}\\ &=\frac{-7}{2\sqrt{-7x - 18}} \end{align*}$$

\]

Answer:

$f(x + h)=\sqrt{-7x-7h - 18}$
$f(x + h)-f(x)=\sqrt{-7x-7h - 18}-\sqrt{-7x - 18}$
$\frac{f(x + h)-f(x)}{h}=\frac{-7}{\sqrt{-7x-7h - 18}+\sqrt{-7x - 18}}$
$f'(x)=\frac{-7}{2\sqrt{-7x - 18}}$