Question

let (f(x)=-7 - 8sqrt{x}). then the expression (\frac{f(x + h)-f(x)}{h}) can be written in the form (\frac{a}{(sqrt{bx+ch})+(sqrt{x})}), where (a), (b), and (c) are constants. (note: its possible for one or more of these constants to be 0.) find the constants.
(a=)
(b=)
(c=)
use your answer from above to find (lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}).
(lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=)
finally, find each of the following:
(f(1)=)
(f(2)=)
(f(3)=)

Explanation:

Step1: Find \(f(x + h)\)

Given \(f(x)=-7 - 8\sqrt{x}\), then \(f(x + h)=-7-8\sqrt{x + h}\)

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

\[

$$\begin{align*} f(x + h)-f(x)&=(-7 - 8\sqrt{x + h})-(-7 - 8\sqrt{x})\\ &=- 8\sqrt{x + h}+8\sqrt{x}\\ &=8(\sqrt{x}-\sqrt{x + h}) \end{align*}$$

\]
Then \(\frac{f(x + h)-f(x)}{h}=\frac{8(\sqrt{x}-\sqrt{x + h})}{h}\)
Rationalize the numerator:
\[

$$\begin{align*} \frac{8(\sqrt{x}-\sqrt{x + h})}{h}&=\frac{8(\sqrt{x}-\sqrt{x + h})(\sqrt{x}+\sqrt{x + h})}{h(\sqrt{x}+\sqrt{x + h})}\\ &=\frac{8(x-(x + h))}{h(\sqrt{x}+\sqrt{x + h})}\\ &=\frac{-8h}{h(\sqrt{x}+\sqrt{x + h})}\\ &=\frac{-8}{\sqrt{x}+\sqrt{x + h}} \end{align*}$$

\]
Comparing with \(\frac{A}{\sqrt{Bx+Ch}+\sqrt{x}}\), we have \(A = - 8\), \(B = 1\), \(C = 1\)

Step3: Find \(\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}\)
\[

$$\begin{align*} \lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h ightarrow0}\frac{-8}{\sqrt{x}+\sqrt{x + h}}\\ &=\frac{-8}{2\sqrt{x}}=-\frac{4}{\sqrt{x}} \end{align*}$$

\]

Step4: Find \(f^{\prime}(1)\), \(f^{\prime}(2)\), \(f^{\prime}(3)\)

When \(x = 1\), \(f^{\prime}(1)=-\frac{4}{\sqrt{1}}=-4\)
When \(x = 2\), \(f^{\prime}(2)=-\frac{4}{\sqrt{2}}=-2\sqrt{2}\)
When \(x = 3\), \(f^{\prime}(3)=-\frac{4}{\sqrt{3}}=-\frac{4\sqrt{3}}{3}\)

Answer:

\(A=-8\)
\(B = 1\)
\(C = 1\)
\(\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=-\frac{4}{\sqrt{x}}\)
\(f^{\prime}(1)=-4\)
\(f^{\prime}(2)=-2\sqrt{2}\)
\(f^{\prime}(3)=-\frac{4\sqrt{3}}{3}\)