Chapter 3.1, Problem 8E

To determine

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

Answer to Problem 8E

The required value is 2.

Explanation of Solution

Given information:

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

Calculation:

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

Write the given definition using the above function as:

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

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

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

Simplify the limit further.

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

Hence, the required value is 2.