Q. 27 - Calculus

Question

Use (a) the $h\to 0$ definition of the derivative and then

(b) the $z\to c$ definition of the derivative to find $f\text{'}\left(c\right)$ for each function f and value $x=c$ in Exercises 23–38.

27. $f\left(x\right)=1-x^3, x=-1$

Step-by-Step Solution

Answer

$f\text{'}(-1)=-3$.

Part (a) Step 1. Given information.

A function is given as $f\left(x\right)=1-x^3$ and $x=c=-1$.

Part (a) Step 2. Solve f ' ( - 1 ) using h → 0 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h} \\ f\text{'}(-1)=\underset{h\to 0}{\lim }\frac{f(-1+h)-f(-1)}{h} \\ =\underset{h\to 0}{\lim }\frac{1-{(h-1)}^3-[1-{(-1)}^3]}{h} \\ =\underset{h\to 0}{\lim }\frac{1-(h^3-1-3h^2+3h)-2}{h} \\ =\underset{h\to 0}{\lim }\frac{1-h^3+1+3h^2-3h-2}{h} \\ =\underset{h\to 0}{\lim }\frac{-h^3+3h^2-3h}{h} \\ =\underset{h\to 0}{\lim }\frac{h(-h^2+3h-3)}{h} \\ =\underset{h\to 0}{\lim }(-h^2+3h-3) \\ =-0+0-3 \\ =-3 \end{gathered}$$

Part (b) Step 1. Solve f ' ( - 1 ) using z → - 1 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c} \\ f\text{'}(-1)=\underset{z\to -1}{\lim }\frac{f\left(z\right)-f(-1)}{z-(-1)} \\ =\underset{z\to -1}{\lim }\frac{f\left(z\right)-f(-1)}{z+1} \\ =\underset{z\to -1}{\lim }\frac{1-z^3-[1-{(-1)}^3]}{z+1} \\ =\underset{z\to -1}{\lim }\frac{1-z^3-2}{z+1} \\ =\underset{z\to -1}{\lim }\frac{-(z^3+1)}{z+1} \\ =\underset{z\to -1}{\lim }\frac{-(z+1)(z^2-z+1)}{z+1} \\ =\underset{z\to -1}{\lim }[-(z^2-z+1\left)\right] \\ =\underset{z\to -1}{\lim }(-z^2+z-1) \\ ={-(-1)}^2+(-1)-1 \\ =-1-2 \\ =-3 \end{gathered}$$