Chapter CR, Problem 1CR

To determine

Find $f\left(x+h\right)\text{ and }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ for the function $f\left(x\right)=x^2-5$ .

Answer to Problem 1CR

The value of $f\left(x+h\right)=x^2+h^2+2hx-5\text{ and }\frac{f\left(x+h\right)-f\left(x\right)}{h}=h+2x$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^2-5$ .

Calculation:

Evaluating $f\left(x+h\right)$ for the function $f\left(x\right)=x^2-5$ by plugging $x=x+h$ in $f\left(x\right)=x^2-5$ .

Therefore,

  $\begin{array}{c}f\left(x+h\right)={\left(x+h\right)}^2-5\\ =x^2+h^2+2hx-5\mathrm{....}\left(1\right)\end{array}$

Similarly, evaluating $\frac{f\left(x+h\right)-f\left(x\right)}{h}$ ,

  $\begin{array}{c}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\left(x^2+h^2+2hx-5\right)-\left(x^2-5\right)}{h}\mathrm{....}\left(\text{From }\left(1\right)\right)\\ =\frac{x^2+h^2+2hx-5-x^2+5}{h}\\ =\frac{h^2+2hx}{h}\\ =h+2x\end{array}$