let (f(x)=-7 - 8sqrt{x}). then the expression (\frac{f(x + h)-f(x)}{h}) can be written in the form…

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.\n(a=)\n(b=)\n(c=)\nuse your answer from above to find (lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}).\n(lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}=)\nfinally, find each of the following:\n(f(1)=)\n(f(2)=)\n(f(3)=)

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.\n(a=)\n(b=)\n(c=)\nuse your answer from above to find (lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}).\n(lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}=)\nfinally, find each of the following:\n(f(1)=)\n(f(2)=)\n(f(3)=)

Answer

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\rightarrow0}\frac{f(x + h)-f(x)}{h})

[ \begin{align*} \lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h\rightarrow0}\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\rightarrow0}\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})