How do you find the local max and min for 2x^3-3x^2-36x-32x3−3x2−36x−3?
1 Answer
Local maximum of
Explanation:
Here's the process, numbers and all:
Find
f(x)=2x^3-3x^2-36x-3
f'(x)=6x^2-6x-36
The local minima and maxima will occur when the derivative equals
6x^2-6x-36=0
x^2-x-6=0
(x-3)(x+2)=0
x=3" ",""" "x=-2
These are the two points at which maxima/minima could occur. There are three ways we can identify which is which:
The First Derivative Test:
Evaluate the sign of the first derivative around each local extremum.
Around
x=-2 :
f'(-3)=36larr"increasing"
f'(-1)=-24larr"decreasing" Since the first derivative switches from increasing to decreasing, there is a local maximum at
x=-2 .Around
x=3 :
f'(2)=-24larr"decreasing"
f'(4)=36larr"increasing" Since the first derivative switches from decreasing to increasing, there is a local minimum at
x=3 .
The Second Derivative Test:
Find the second derivative of the function. Then, find the sign of it at each point.
-
If
f''(a)<0 andf'(a)=0 , then there is a local maximum atx=a . -
If
f''(a)>0 andf'(a)=0 , then there is a local minimum atx=a .
f'(x)=6x^2-6x-36
f''(x)=12x-6
Determine the sign at each extremum.
f''(-2)=-30 Since this is
<0 , there is a local maximum atx=-2 .
f''(3)=30 Since this is
>0 , there is a local minimum atx=3 .
Check a graph of the original function:
This method shouldn't be used as proof on a test, say, but it's a fine way to make sure you're on the right track with your answer.
graph{2x^3-3x^2-36x-3 [-5, 7, -120, 150]}
Indeed, there is a local maximum at
Thus, the local maximum is