use the function below to answer parts (a)-(c).\n$f(x)=\\frac{2}{x}$\n(a) use the formal definition to find…

use the function below to answer parts (a)-(c).\n$f(x)=\\frac{2}{x}$\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 5$.\n(b) find $f(5)$ and find the equation of the normal line at the point $(5, f(5))$.\n(c) graph $y = f(x)$ and the tangent line at the point $(5, f(5))$ in the same coordinate system.\n(a) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=\\lim_{h\\to0}\\frac{f(x + h)-f(x)}{h}$ provided that the limit exists. use the definition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 5$.\n$f(5)=\\lim_{h\\to0}\\frac{\\square-\\left\\frac{2}{x}\\right}{h}$

use the function below to answer parts (a)-(c).\n$f(x)=\\frac{2}{x}$\n(a) use the formal definition to find the derivative of $y = f(x)$ at $x = 5$.\n(b) find $f(5)$ and find the equation of the normal line at the point $(5, f(5))$.\n(c) graph $y = f(x)$ and the tangent line at the point $(5, f(5))$ in the same coordinate system.\n(a) the derivative of a function $f$ at $x$, denoted by $f(x)$, is $f(x)=\\lim_{h\\to0}\\frac{f(x + h)-f(x)}{h}$ provided that the limit exists. use the definition of the derivative of $f$ at $x$ to find the derivative of the given function $f(x)$ when $x = 5$.\n$f(5)=\\lim_{h\\to0}\\frac{\\square-\\left\\frac{2}{x}\\right}{h}$

Answer

Explanation:

Step1: Substitute (f(x)) and (f(x + h)) into the derivative formula

Given (f(x)=\frac{2}{x}), then (f(5)=\frac{2}{5}) and (f(5 + h)=\frac{2}{5 + h}). [ f^{\prime}(5)=\lim_{h\rightarrow0}\frac{\frac{2}{5 + h}-\frac{2}{5}}{h} ]

Step2: Simplify the numerator

[ \begin{align*} \frac{2}{5 + h}-\frac{2}{5}&=\frac{2\times5-2\times(5 + h)}{5(5 + h)}\ &=\frac{10-10 - 2h}{5(5 + h)}\ &=\frac{-2h}{5(5 + h)} \end{align*} ] So (f^{\prime}(5)=\lim_{h\rightarrow0}\frac{\frac{-2h}{5(5 + h)}}{h})

Step3: Simplify the fraction

[ \begin{align*} \frac{\frac{-2h}{5(5 + h)}}{h}&=\frac{-2h}{5(5 + h)}\times\frac{1}{h}\ &=\frac{-2}{5(5 + h)} \end{align*} ]

Step4: Evaluate the limit

[ \begin{align*} f^{\prime}(5)&=\lim_{h\rightarrow0}\frac{-2}{5(5 + h)}\ &=\frac{-2}{5\times(5+0)}\ &=-\frac{2}{25} \end{align*} ]

Answer:

(f^{\prime}(5)=-\frac{2}{25})