Chapter 3.4, Problem 42E

To determine

To find:The instantaneous rate of change of the function.

Answer to Problem 42E

Correct option that represents instantaneous rate of change of given function at $x=-1$ is C that is $0$ .

Explanation of Solution

Given information:

1. Function for which instantaneous rate of change needs to be determine is as follows

$f\left(x\right)=x^2-\frac{2}{x}+4$ .

2. Instantaneous rate of change needs to be determine at $x=-1$ .
3. Five given options of rate change are as follows

1. $-7$
2. $-4$
3. $0$
4. $4$
5. $7$

Formula used:

1. Instantaneous rate of change of any function $f\left(x\right)$ at $x=a$ is defined by $f^{\prime }\left(a\right)$using following formula.

  $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f(a)}{h}$

Calculation:

The Function for which Instantaneous rate of change needs to be determine is as follows

  $f\left(x\right)=x^2-\frac{2}{x}+4$

Instantaneous rate of change of $f\left(x\right)$ at $x=-1$ is defined by $f^{\prime }\left(-1\right)$ .

Evaluate $f^{\prime }\left(x\right)$ .

  $f^{\prime }\left(-1\right)=\underset{h\to 0}{\lim }\frac{f\left(-1+h\right)-f(-1)}{h}$

  $=\underset{h\to 0}{\lim }\frac{\left({\left(-1+h\right)}^2-\frac{2}{\left(-1+h\right)}+4\right)-\left({\left(-1\right)}^2-\frac{2}{\left(-1\right)}+4\right)}{h}$

  $=\underset{h\to 0}{\lim }\frac{\left(1^2+h^2-2h-\frac{2}{\left(h-1\right)}+4\right)-\left(1+2+4\right)}{h}$

  $=\underset{h\to 0}{\lim }\frac{h^2-2h-\frac{2}{\left(h-1\right)}+5-7}{h}$

  $=\underset{h\to 0}{\lim }\frac{\left(h^2-2h\right)\left(h-1\right)-2-2\left(h-1\right)}{h\left(h-1\right)}$

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

  $=\underset{h\to 0}{\lim }\frac{h^3-3h^2}{h\left(h-1\right)}$

  $=\underset{h\to 0}{\lim }\frac{\left(h^2-3h\right)}{\left(h-1\right)}$

  $=\frac{0}{-1}$

  $=0$

Hence Instantaneous rate of change of $f\left(x\right)=x^2-\frac{2}{x}+4$ at $x=-1$ is $0$ that means correct option is C.