let $f(x)=sqrt{17 - x}$. compute $f(8)$ using the limit definition
let $f(x)=sqrt{17 - x}$. compute $f(8)$ using the limit definition
Answer
Explanation:
Step1: Recall the limit - definition of the derivative
The limit - definition of the derivative of a function $y = f(x)$ is $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. For $f(x)=\sqrt{17 - x}$, we have $f(x + h)=\sqrt{17-(x + h)}$.
Step2: Substitute into the limit - definition
[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{\sqrt{17-(x + h)}-\sqrt{17 - x}}{h}\ &=\lim_{h\rightarrow0}\frac{\sqrt{17-(x + h)}-\sqrt{17 - x}}{h}\cdot\frac{\sqrt{17-(x + h)}+\sqrt{17 - x}}{\sqrt{17-(x + h)}+\sqrt{17 - x}}\ &=\lim_{h\rightarrow0}\frac{(17-(x + h))-(17 - x)}{h(\sqrt{17-(x + h)}+\sqrt{17 - x})}\ &=\lim_{h\rightarrow0}\frac{17-x - h-17 + x}{h(\sqrt{17-(x + h)}+\sqrt{17 - x})}\ &=\lim_{h\rightarrow0}\frac{-h}{h(\sqrt{17-(x + h)}+\sqrt{17 - x})}\ &=\lim_{h\rightarrow0}\frac{-1}{\sqrt{17-(x + h)}+\sqrt{17 - x}} \end{align*} ]
Step3: Evaluate the limit
As $h\rightarrow0$, we get $f^{\prime}(x)=\frac{-1}{2\sqrt{17 - x}}$.
Step4: Find $f^{\prime}(8)$
Substitute $x = 8$ into $f^{\prime}(x)$. Then $f^{\prime}(8)=\frac{-1}{2\sqrt{17 - 8}}=\frac{-1}{2\sqrt{9}}=\frac{-1}{6}$.
Answer:
$-\frac{1}{6}$