What are the absolute extrema of # f(x)= xsqrt(25-x^2) in [-4,5]#?

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Answer 1 · 2016-02-05T16:17:04

The absolute minimum is #-25/2# (at #x=-sqrt(25/2)#). The absolute maximum is #25/2# (at #x=sqrt(25/2)#).

#f(-4) = -12# and #f(5)=0#

#f'(x) = sqrt(25-x^2)+x/(cancel(2)sqrt(25-x^2))*-cancel(2)x#

# = (25-x^2-x^2)/sqrt(25-x^2) = (25-2x^2)/sqrt(25-x^2)#

The critical numbers of #f# are #x=+-sqrt(25/2)# Both of these are in #[-4,5]#..

#f(-sqrt(25/2)) = -sqrt(25/2)sqrt(25-25/2)#

# = -sqrt(25/2)sqrt(25/2) = -25/2#

By symmetry (#f# is odd), #f(sqrt(25/2)) = 25/2#

Summary:
#f(-4) = -12#
#f(-sqrt(25/2)) = -25/2#
#f(sqrt(25/2)) = 25/2#
#f(5)=0#

The absolute minimum is #-25/2# (at #x=-sqrt(25/2)#).
The absolute maximum is #25/2# (at #x=sqrt(25/2)#).