Question

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

Answer 1

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]}