use the function below to answer parts (a)-(c).\nf(x)=\\frac{4}{x}\n(a) use the formal definition to find…

use the function below to answer parts (a)-(c).\nf(x)=\\frac{4}{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.\nf(5)=\\lim_{h\\to0}\\frac{\\square-\\left\\frac{4}{x}\\right}{h}

use the function below to answer parts (a)-(c).\nf(x)=\\frac{4}{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.\nf(5)=\\lim_{h\\to0}\\frac{\\square-\\left\\frac{4}{x}\\right}{h}

Answer

Explanation:

Step1: Substitute ( f(x)=\frac{4}{x} ) into the derivative formula

[ \begin{align*} f^{\prime}(5)&=\lim_{h\rightarrow0}\frac{f(5 + h)-f(5)}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{4}{5 + h}-\frac{4}{5}}{h} \end{align*} ]

Step2: Simplify the numerator

[ \begin{align*} \frac{4}{5 + h}-\frac{4}{5}&=\frac{4\times5-4\times(5 + h)}{5(5 + h)}\ &=\frac{20-20-4h}{5(5 + h)}\ &=\frac{-4h}{5(5 + h)} \end{align*} ]

Step3: Substitute the simplified numerator back into the limit

[ \begin{align*} f^{\prime}(5)&=\lim_{h\rightarrow0}\frac{\frac{-4h}{5(5 + h)}}{h}\ &=\lim_{h\rightarrow0}\frac{-4h}{5h(5 + h)}\ &=\lim_{h\rightarrow0}\frac{-4}{5(5 + h)} \end{align*} ]

Step4: Evaluate the limit

As ( h\rightarrow0 ), we substitute ( h = 0 ) into ( \frac{-4}{5(5 + h)} ) [ f^{\prime}(5)=\frac{-4}{5\times(5+0)}=-\frac{4}{25} ]

Answer:

( f^{\prime}(5)=-\frac{4}{25} )