Find points at which the derivative of a given function has specified values

by RoRi

September 11, 2015

Let

$\frac{1}{3} x^3 + \frac{1}{2} x^2 -2x.$

Find values such that,

Using the formula for the derivative of a polynomial we first compute

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

Then,

We have
$f'(x) = 0 \quad \implies \quad x^2 + x - 2 = 0 \quad \implies \quad x = \{1, -2 \}.$
We have
$f'(x) = -2 \quad \implies \quad x^2 + x - 2 = -2 \quad \implies \quad x = \{0, -1 \}.$
We have
$f'(x) = 10 \quad \implies \quad x^2 + x - 2 = 10 \quad \implies \quad x = \{ 3, -4 \}.$

Stumbling Robot