What are the extrema of #f(x)=-8x^2+x# on #[-4,8]#?

1 Answer
Feb 9, 2016

Answer:

Absolute minimum of #-512# at #x=8# and an absolute maximum of #1/32# at #x=1/16#

Explanation:

When finding the extrema on an interval, there are two locations they could be: at a critical value, or at one of the endpoints of the interval.

To find the critical values, find the function's derivative and set it equal to #0#. Since #f(x)=-8x^2+x#, through the power rule we know that #f'(x)=-16x+1#. Setting this equal to #0# leaves us with one critical value at #x=1/16#.

Thus, our locations for potential maxima and minima are at #x=-4#, #x=1/16#, and #x=8#. Find each of their function values:

#f(-4)=-8(-4)^2-4=ul(-132)#

#f(1/16)=-8(1/16)^2+1/16=-1/32+1/16=ul(1/32)#

#f(8)=-8(8)^2+8=ul(-504)#

Since the highest value is #1/32#, this is the absolute maximum on the interval. Note that the maximum itself is #1/32#, but its location is at #x=1/16#. Likewise, the lowest value and absolute minimum is #-512#, located at #x=8#.

This is #f(x)# graphed: you can see that its maxima and minima are indeed where we found.

graph{-8x^2+x [-4.1, 8.1, -550, 50]}