Inactive Tutor answered • 08/11/15
Tutor
New to Wyzant
Hi, Jasmyn!
The number of local maximums and minimums of a function are calculated by determining how many places have a slope of zero. In other words, you need to answer the question "how many zeroes does the function f'(x) have?"
In this case, you don't need to know where the maxs and mins are, so you can just evaluate how many distinct zeroes show up in f'(x):
f'(x) = x * (x-3)2 * (x+1)4
. . . x = 0, (x-3)2=0 , (x+1)4=0
There are 3 distinct possibilities where f'(x)=0 (located at x=0, x=3, and x=-1).
I hope this helps!
~Patty
Inactive Tutor
Always use test points to ensure that you do get a max or min at those x values. For example:
if f'(-1) is negative, and f'(1) is also negative, then x=0 has neither a maximum nor a minimum. But if
f'(-1) is negative, and f'(1) is positive, then x=0 has a minimum value.
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08/11/15
Jasmyn G.
Thank you so much! This really helped and explained in vivid detail! Much appreciated!!
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08/11/15
Inactive Tutor
If you use the test point approach, be sure to use test points that lie within the interval defined by adjacent zeros of the derivative. Example, for x=0, the adjacent zeros of f'(x) are x=-1 and x=3, so your test points should be chosen from the open intervals (-1,0) and (0,3).
Sometimes, a simpler approach is to check the second derivative. If f'(x)=0, then there is a minimum if f"(x)>0 and a maximum if f"(x)<0. However, you can not be sure if f"(x)=0. In that case you can go back to the test points.
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08/12/15
Mark M.
08/11/15