1. If f'(x)=x^3-12x+2, at what values of x would f ‘(x) have roots?
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This looks like a very nice straightforward question. Of course, you still have to know what all the words mean!
i.e. at what values would put the polynomial =0?
So let f'(x)=0 and solve for x.
These are the x-coordinates hitting the x-axis.
f'(x)=3x^2-12
Setting f(x) equal to o:
3x^2-12=0
(3x-6)(x+2)=0
and
(3x+6)(x-2)=0
Both ways x= -2 or 2 These are f'(x)’s roots.
Oops. Line 2 should read:
Setting f'(x) equal to 0: