Question

Find the absolute extrema of the function f(x) = (x2 + 8)"3 on the interval - 1,5). The absolute maximum occurs at x = (Simplify your answer. Use a comma to separate answers as needed.) The absolute minimum occurs at x = (Simplify your answer. Use a comma to separate answers as needed.)

Answer 1

Given f(x)=(x2+8) Y3 To find absolute extrema of f(x) on [-1, s] plug in all Critical numbers and of interval in f(x), Highest value them results absolute Maximum and value results absolute end points among least Minimum. am nong them, Critical numbers: flow=0 413 fica ex (x2 +8) fa (an) = nan- = } (x2 +8'))") da (258) =% (+8)+73 (272='+0) CS Scanned with CamScanner

22 if'(x)=0 1 ! F'(x)= 3(x2+8143 22 FO S 2750 x=0 3(x2+8) 13 EN Y3 (9) た 73 2 (2 critical number is x=0 plug in a="1, 0,5 in f(x) = 68+8) } *==FC-1) = (6)2+8)Y? 21=0.5) f(0) = (0+8) Y3 23) 7455 =) f(s) = (15)2+8795-(33) highest occurs at x=5 value V3 is 3 manimum. absolute rescelts at n=0 results occurs value least is 2 at x=15 minimum. occurs absolute mademum X= 0 at occurs .. The absolute Absolute minimum CS Scanned with CamScanner