Question

a. Find the open interval(s) on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur.

Answer 1

The given function is f(1) = x3 (2+ 24) f'() = *** (1) + 4623943 (*+84) A = 30 +H+ 24 3 (3) 43 45+24 - 3 (22) 3/3 To find critical points solve f'(x)=0 for the > f (2) = 44724 3 (21) 43 to → 44+24 50 441 = -94

H=-6 nes the only caifical point is at r=-6 - from 0 f'(X) = 4 (3+6) 3(74)213 Clear any f'() >0 face HE (-600) 0% function is increasing on 6-6600) and since t can) <o on o function is decreasing (-00,-6) on 600,-6) db) foco

= 23 (4) – (4 *++)( * tvo) 3(823)2 = 4(3)# -(4x+84).2 3443 3 (22) 7/3 124-84-48 9 (7)S/3 f" (a)= 1 44-48 9 (26) 5/3 At r=-6, = -24 - 48 q (-6353 . _-ve =tre -vre 68 At H=-6, function has local minema.

And local menimum walele is given .. by: f66)= 66% (-6724) = 1966) = -18 (6) Ars. and function has no local maxima.