Chapter 14.4, Problem 21AYU

To determine

To find: The derivative of $f\left(x\right) = -4x + 5$ at 3.

Answer to Problem 21AYU

$-4$

Explanation of Solution

Given:

$f\left(x\right) = -4x + 5$

$c = 3$

Formula used:

The derivative of $y = f\left(x\right)$ at $c$ is given by:

$f^{\prime }\left(c\right) = \underset{x\to c}{\lim }\frac{f\left(x\right) - f\left(c\right)}{x - c}$

Calculation:

Here, $f\left(x\right) = -4x + 5$ and $c = 3$.

$f\left(c\right) = f\left(3\right) = -4\left(3\right) + 5 = -12 + 5 = -7$

Therefore,

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

$f^{\prime }\left(3\right) = \underset{x\to 3}{\lim }\left[\frac{-4x + 5 - \left(-7\right)}{x - 3}\right]$

$f^{\prime }\left(3\right) = \underset{x\to 3}{\lim }\left(\frac{-4x + 5 + 7}{x - 3}\right)$

$f^{\prime }\left(3\right) = \underset{x\to 3}{\lim }\left(\frac{-4x + 12}{x - 3}\right)$

$f^{\prime }\left(3\right) = \underset{x\to 3}{\lim }\left[-\frac{4\left(x - 3\right)}{\left(x - 3\right)}\right]$

$f^{\prime }\left(3\right) = \underset{x\to 3}{\lim }\left(-4\right)$

$f^{\prime }\left(3\right) = -4$