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

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Answer 1 · 2016-01-06T15:42:02

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

#f(x)=2x^3-15x^2# is continuous on the closed interval #[-2,10]#, 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) = 6x^2-30x = 6x(x-5)#.

#f'# is never undefined and #f'(x)=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(-2) = -16-25(4) = -16-60 = -76#

#f(0) = 0#

#f(5) = 2(5)^3-15(5)^2 = 10(25)-15(25) = -5(25)=-125#

#f(10) = 2000-1500 = 500#

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