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#.