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

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Answer 1 · 2016-01-06T15:56:34

The minimum is $39/8$ and the maximum is $50$ .

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

$f'(x) = 4x-1$ .

$f'$ is never undefined and $f'(x)=0$ at $x=1/4$ .

Both $1/4$ is in the interval of interest (the domain of this problem).

Do the arithmetic to find

$f(-1) =2+1+5=8$

$f(1/4) = 2/16-1/4+5 = 1/8-2/8+40/8 = 39/8$

$f(5) = 50-5+5=50$

The minimum is $39/8$ and the maximum is $50$ .

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