# What are the absolute extrema of f(x)=2x^3-15x^2 in[-2,10]?

Jan 6, 2016

Minimum: $- 125$, Maximum: $500$.

#### Explanation:

$f \left(x\right) = 2 {x}^{3} - 15 {x}^{2}$ is continuous on the closed interval $\left[- 2 , 10\right]$, so it has a minimum and a maximum on the interval. (Extreme Value Theorem)

The extrema must occur at either an endpoint of the interval or at a critical number for $f$ in the interval.

Find the critical numbers for $f$.

$f ' \left(x\right) = 6 {x}^{2} - 30 x = 6 x \left(x - 5\right)$.

$f '$ is never undefined and $f ' \left(x\right) = 0$ at $x = 0$ and $5$.

Both $0$ and $5$ are in the interval of interest (the domain of this problem).

Do the arithmetic to find

$f \left(- 2\right) = - 16 - 25 \left(4\right) = - 16 - 60 = - 76$

$f \left(0\right) = 0$

$f \left(5\right) = 2 {\left(5\right)}^{3} - 15 {\left(5\right)}^{2} = 10 \left(25\right) - 15 \left(25\right) = - 5 \left(25\right) = - 125$

$f \left(10\right) = 2000 - 1500 = 500$

Minimum: $- 125$, Maximum: $500$.