Question

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

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 \(f^{\prime}(4)=\lim_{h
ightarrow0}\frac{\frac{1}{4 + h}-\frac{1}{4}}{h}\).

Step2: Simplify the numerator

$$
LATEXBLOCK0
$$

So \(f^{\prime}(4)=\lim_{h
ightarrow0}\frac{\frac{-h}{4(4 + h)}}{h}\).

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 \(h
eq0\)).

Step4: Evaluate the limit

\(\lim_{h
ightarrow0}-\frac{1}{4(4 + h)}=-\frac{1}{16}\)

Answer:

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