What are the critical values, if any, of f(x) = x^3 + x^2 - x ?

Mar 3, 2018

Critical points: $\left(0.43 , - 0.49\right)$ and $\left(- 0.77 , - 2.23\right)$

Explanation:

We are given: $f \left(x\right) = {x}^{3} + {x}^{2} - x$

We compute the derivative:
$f ' \left(x\right) = 3 {x}^{2} + x - 1$

We set the derivative to zero to find critical points:
$f ' \left(x\right) = 0$
$3 {x}^{2} + x - 1 = 0$

This cannot be factored and solved. We need to use the quadratic equation:

$x = \frac{- b + = \sqrt{{b}^{2} - 4 a c}}{2 a}$
where $a = 3$, $b = 1$, and $c = - 1$.

This gives:
$x = 0.43$ and $x = - 0.77$

Substituting these values into $f \left(x\right)$ gives:
$f \left(0.43\right) = - 0.49$
$f \left(- 0.77\right) = - 2.23$

Hence the critical points are:
$\left(0.43 , - 0.49\right)$ and $\left(- 0.77 , - 2.23\right)$