f(x)=\\frac{1}{x}\n(a) use the formal definition to find the derivative of y = f(x) at x = 4.\n(b) find f(4)…

f(x)=\\frac{1}{x}\n(a) use the formal definition to find the derivative of y = f(x) at x = 4.\n(b) find f(4) and find the equation of the normal line at the point (4, f(4)).\n(c) graph y = f(x) and the tangent line at the point (4, f(4)) 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 = 4.\nf(4)=\\lim_{h\\to0}\\frac{\\square-\\left\\frac{1}{x}\\right}{h}

f(x)=\\frac{1}{x}\n(a) use the formal definition to find the derivative of y = f(x) at x = 4.\n(b) find f(4) and find the equation of the normal line at the point (4, f(4)).\n(c) graph y = f(x) and the tangent line at the point (4, f(4)) 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 = 4.\nf(4)=\\lim_{h\\to0}\\frac{\\square-\\left\\frac{1}{x}\\right}{h}

Answer

Explanation:

Step1: Substitute (x = 4) into the formula

We know (f(x)=\frac{1}{x}), so (f(4 + h)=\frac{1}{4 + h}) and (f(4)=\frac{1}{4}). Then (f^{\prime}(4)=\lim_{h\rightarrow0}\frac{\frac{1}{4 + h}-\frac{1}{4}}{h}).

Step2: Simplify the numerator

[ \begin{align*} \frac{1}{4 + h}-\frac{1}{4}&=\frac{4-(4 + h)}{4(4 + h)}\ &=\frac{4-4 - h}{4(4 + h)}\ &=\frac{-h}{4(4 + h)} \end{align*} ] So (f^{\prime}(4)=\lim_{h\rightarrow0}\frac{\frac{-h}{4(4 + h)}}{h}).

Step3: Simplify the fraction

(\frac{\frac{-h}{4(4 + h)}}{h}=\frac{-h}{4(4 + h)}\times\frac{1}{h}=-\frac{1}{4(4 + h)}) (for (h\neq0)).

Step4: Evaluate the limit

(\lim_{h\rightarrow0}-\frac{1}{4(4 + h)}=-\frac{1}{16})

Answer:

(f^{\prime}(4)=-\frac{1}{16})