Question

Use (a) the h→0 definition of the derivative and then (b) the z→c definition of the derivative to find f'(c) for each function f and value x = c.

$f\left(x\right)=\frac{x^2-3x}{x+1},x=0$

Short answer

(a) $f^{\text{'}}\left(c\right)=-3$

(b)$f^{\text{'}}\left(c\right)=-3$

Step-by-step solution

Part (a) Step 1. Given information.

Given function is $f\left(x\right)=\frac{x^2-3x}{x+1}$

We have to find$f^{\text{'}}\left(c\right)$at$x=0$

Part (a) Step 2. Find the f'(c)

We have to find the derivative of the function using h→0 definition,

Therefore,

$\begin{array}{r}\underset{h\to 0}{lim} \frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim} \frac{\frac{(0+h{)}^2-3(0+h)}{0+h+1}-\frac{0-0}{0+1}}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{h^2-3h}{h+1}}{h}\\ =\underset{h\to 0}{lim} \frac{h(h-3)}{h(h+1)}\\ =\underset{h\to 0}{lim} \frac{h(h-3)}{h(h+1)}\\ =\underset{h\to 0}{lim} \frac{h-3}{h+1}\\ =-3\end{array}$

Part (b) Step 1. Find f'(c)

Find the derivate of the function using $x\to 0$definition,

$\begin{array}{r}\underset{x\to 0}{lim} \frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{lim} \frac{\frac{x^2-3x}{x+1}-\frac{0-0}{0+1}}{x}\\ =\underset{x\to 0}{lim} \frac{x(x-3)}{x(x+1)}\\ =\underset{x\to 0}{lim} \frac{x-3}{x+1}\\ =-3\end{array}$