Find points of a function at which the tangent line has specified values

by RoRi

September 16, 2015

Consider the function

$f(x) = \frac{2}{3}x^3 + \frac{1}{2}x^2 -x-1.$

Find the points at which the slope of $f$ is

The derivative is given by

$f'(x) = 2x^2 + x - 1 = (2x-1)(x+1).$

Then,

The slope is 0 means
$f'(x) = 0 \ \implies \ (2x-1)(x+1) = 0 \ \implies \ x = \left\{ -1, \frac{1}{2} \right\}.$
The slope is -1 means
$f'(x) = -1 \ \implies \ 2x^2 + x - 1 = -1 \ \implies \ 2x^2 + x = 0 \ \implies \ \left\{ 0, -\frac{1}{2} \right\}.$
The slope is 5 means
$f'(x) = 5 \ \implies \ 2x^2 + x - 1 = 5 \ \implies \ 2x^2 + x - 6 = 0 \ \implies \ x = \left\{ -2, \frac{3}{2} \right\}.$

Stumbling Robot