Chapter 3.1, Problem 3E

To determine

To calculate: The derivative of the given function at the given value of a using the definition $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Answer to Problem 3E

The required value is 2.

Explanation of Solution

Given information:

The function: $f\left(x\right)=3-x^2$ , $a=-1$

Calculation:

The given function is $f\left(x\right)=3-x^2$ .

Write the given definition using the above function as:

  $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\left[\frac{\left\{3-{\left(a+h\right)}^2\right\}-\left(3-a^2\right)}{h}\right]$

Substitute $-1$ for a in the above limit and simplify.

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

Simplify the limit further.

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

Substitute the 0 for h in the limit and simplify.

  $\begin{array}{c}{f}^{\prime }\left(-1\right)=2-0\\ =2\end{array}$

Hence, the required value is 2.