let (f(x)=sqrt{x + 6}).\na. determine (f(x + h)).\n(f(x + h)=)\n\nb. determine (f(x + h)-f(x)).\n(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\rightarrow0) of (\frac{f(x + h)-f(x)}{h}). (f'(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h\rightarrow0}\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}})