What are the extrema of $x^{3} - 2 x^{2} - 8 x$ $x^3-2x^2-8x$ ?

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Answer 1 · 2015-12-23T15:32:33

$\frac{2 + 2 \sqrt{7}}{3} and \frac{2 - 2 \sqrt{7}}{3}$ $[2+ 2sqrt(7)]/3 and [2- 2sqrt(7)]/3$

$f (x) = x^{3} - 2 x^{2} - 8 x$ $f(x)=x^3 - 2x^2 - 8x$
$f'(x)= 3x^2 - 4x -8$
$3x^2 - 4x -8=0$

since this is not factorable we use the quadratic equation

$x= [2+ 2sqrt(7)]/3$

What are the extrema of x^3-2x^2-8xx3−2x2−8xx^3-2x^2-8x?