How do you find the local max and min for # 2x^3-3x^2-36x-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# and#f'(a)=0# , then there is a local maximum at#x=a# . -
If
#f''(a)>0# and#f'(a)=0# , then there is a local minimum at#x=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 at#x=-2# .
#f''(3)=30# Since this is
#>0# , there is a local minimum at#x=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