Chapter 14.4, Problem 29AYU

To determine

To find: The derivative of $f\left(x\right) = x^3 + x^2 - 2x$ at 1.

Answer to Problem 29AYU

3

Explanation of Solution

Given:

$f\left(x\right) = x^3 + x^2 - 2x$

$c = 1$

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, $f\left(x\right) = x^3 + x^2 - 2x$ and $c = 1$.

$f\left(c\right) = f\left(1\right) = 1^3 + 1^2 - 2\left(1\right) = 2 - 2=0$

Therefore,

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

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

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

$f^{\prime }\left(1\right) = \underset{x\to 1}{\lim }\left[\frac{x\left(x^2 + x - 2\right)}{\left(x - 1\right)}\right]$

$f^{\prime }\left(1\right) = \underset{x\to 1}{\lim }\left[\frac{x\left(x + 2\right)\left(x - 1\right)}{\left(x - 1\right)}\right]$

$f^{\prime }\left(1\right) = \underset{x\to 1}{\lim }\left[x\left(x + 2\right)\right]$

$f^{\prime }\left(1\right) = 1\left(1 + 2\right)$

$f^{\prime }\left(1\right) = 3$