Question

let (f(x)=sqrt{x + 6}).
a. determine (f(x + h)).
(f(x + h)=)

b. determine (f(x + h)-f(x)).
(f(x + h)-f(x)=)

c. determine (\frac{f(x + h)-f(x)}{h}).
(\frac{f(x + h)-f(x)}{h}=)

d. determine (f(x)).
(f(x)=)

Explanation:

Step1: Find f(x + h)

Substitute \(x+h\) into \(f(x)\). So \(f(x + h)=\sqrt{(x + h)+6}=\sqrt{x + h+6}\)

Step2: Find f(x + h)-f(x)

\(f(x + h)-f(x)=\sqrt{x + h+6}-\sqrt{x + 6}\)

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

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{\sqrt{x + h+6}-\sqrt{x + 6}}{h}\\ &=\frac{(\sqrt{x + h+6}-\sqrt{x + 6})(\sqrt{x + h+6}+\sqrt{x + 6})}{h(\sqrt{x + h+6}+\sqrt{x + 6})}\\ &=\frac{(x + h+6)-(x + 6)}{h(\sqrt{x + h+6}+\sqrt{x + 6})}\\ &=\frac{x + h+6 - x-6}{h(\sqrt{x + h+6}+\sqrt{x + 6})}\\ &=\frac{h}{h(\sqrt{x + h+6}+\sqrt{x + 6})}\\ &=\frac{1}{\sqrt{x + h+6}+\sqrt{x + 6}} \end{align*}$$

\]

Step4: Find \(f'(x)\)

Take the limit as \(h
ightarrow0\) of \(\frac{f(x + h)-f(x)}{h}\).
\(f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{1}{\sqrt{x + h+6}+\sqrt{x + 6}}=\frac{1}{2\sqrt{x + 6}}\)

Answer:

a. \(\sqrt{x + h+6}\)
b. \(\sqrt{x + h+6}-\sqrt{x + 6}\)
c. \(\frac{1}{\sqrt{x + h+6}+\sqrt{x + 6}}\)
d. \(\frac{1}{2\sqrt{x + 6}}\)