Relative extrema

Please find the solution in the attachment.

Find all relative extrema of the following function:

f(x) = x^3 - 6x^2 +15
Solution:
To find the relative extrema, we need to first differentiate the function.
So, differentiating f(x) with respect to x, we get f'(x)=3x2-12x
As the next step, we make f'(x) equal to zero. Then we get 3x2-12x=0. So, we get x=0 or 4
Next, we differentiate further to get f"(x)=6x-12
Now we have f"(0)=-12 <0. Hence the function f(x) has relative maximum at x=0
Further, we have f"(4)=12>0. Hence the function f(x) has relative minimum at x=4.
Hence the relative extrema of f(x) are given by f(0)=15 (relative maximum) and f(4)=-17 (relative minimum).

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