The derivative of a function f is given by f'(x) = (x^3-2x)(cosx) for 0≤x≤2.
a) Find the x-coordinate of the relative minimum of f(x). Justify your answer.
b) Find the x-coordinate of any points of inflection of the graph of f(x) on the given interval. Justify your answer.

Question

The derivative of a function f is given by f'(x) = (x^3-2x)(cosx) for 0≤x≤2.
a) Find the x-coordinate of the relative minimum of f(x). Justify your answer.
b) Find the x-coordinate of any points of inflection of the graph of f(x) on the given interval. Justify your answer.

ranga · Answer

Equate f'(x) to zero, solve for x.  That will give your critical numbers.
Find f''(x) by differentiating f'(x) once.
Substitute the critical numbers and see which one gives a positive value for x in the range [0,2]

jaredstone4 · Answer

Okay, I got the critical numbers to be x = 0, ±sqrt(2), and pi/2.

ranga · Answer

Yes, but not -sqrt(2) since it is outside the domain of [0,2].

Find f''(x) and test each of the three critical numbers to see which gives f''(x) > 0.

jaredstone4 · Answer

Oh okay. And then how would you justify it? The explanations are always what confused me.

ranga · Answer

Justification is: If f'(a) is zero and f''(a) is positive then x = a is a local minimum.

jaredstone4 · Answer

Great, thanks so much. For part (b), I just set f"(x) = 0 right?

ranga · Answer

Yes.

ranga · Answer

You are welcome.

jaredstone4 · Answer

As for justification, is it just that whatever x-coordinate I find is a point of inflection because f"(x) =0? Or do I need to mention something about changing concavity...