# What are the absolute extrema of  f(x)= 2x^2 - x +5 in [-1, 5]?

Jan 6, 2016

The minimum is $\frac{39}{8}$ and the maximum is $50$.

#### Explanation:

$f \left(x\right) = 2 {x}^{2} - x + 5$ is continuous on the closed interval $\left[- 1 , 5\right]$, 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 $f$.

$f ' \left(x\right) = 4 x - 1$.

$f '$ is never undefined and $f ' \left(x\right) = 0$ at $x = \frac{1}{4}$.

Both $\frac{1}{4}$ is in the interval of interest (the domain of this problem).

Do the arithmetic to find

$f \left(- 1\right) = 2 + 1 + 5 = 8$

$f \left(\frac{1}{4}\right) = \frac{2}{16} - \frac{1}{4} + 5 = \frac{1}{8} - \frac{2}{8} + \frac{40}{8} = \frac{39}{8}$

$f \left(5\right) = 50 - 5 + 5 = 50$

The minimum is $\frac{39}{8}$ and the maximum is $50$.