What are the extrema of #f(x)=2x^3-3x^2-36x-3#?

1 Answer
Sep 8, 2017

Answer:

Local maximam is at # (-2,41)# and local minimum
is at
# (3,-84)#

Explanation:

All local maximums and minimums on a function’s graph are called local extrema.

#f(x) =2x^3-3x^2-36x-3 or f^'(x) = 6x^2 -6x- 36#

At critical points #f^'(x)=0 :. 6(x^2-x-6)=0 # or

#x^2-x-6=0 or (x-3)(x+2) =0 # ;Critical points are

#x=-2, x=3 # when #x=-2 , f(x)= 2(-2)^3-3(-2)^2-36(-2)-3=41#

when #x=3 , f(x)= 2* 3^3-3*3^2-36*3-3= -84# . To test

local maximum or minimum we will examine sign change

at three positions #x = -3 , x =0 ,x=4 # for

#f^'(x) = 6x^2 -6x- 36 ; f^'( -3) = 36 # (increasing) ,

#f^'(0) = -36 # (decreasing) and #f^'(4) = 36 # (increasing)

#x=-2# is local maximum and #x=3 # is local minimum

Hence local maximam is at # (-2,41)# and local minimum

is at # (3,-84)#

graph{2x^3-3x^2-36x-3 [-157, 157, -78.5, 78.5]}