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

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Answer 1 · 2018-03-03T19:36:49

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

We are given: $f(x) = x^3 +x^2-x$

We compute the derivative:
$f'(x) = 3x^2+x-1$

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

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

$x = (-b += sqrt(b^2-4ac))/(2a)$
where $a = 3$ , $b=1$ , and $c=-1$ .

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

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

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

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