Question

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 3x(1/3) + 6x(4/3). You must justify your answer using an analysis of f '(x) and f "(x).

My start to a solution:
x-values: (-1/8) and (-1/4)

I got -1/8 from f'(x)- extrema
I got -1/4 from f"(x)- inflection point

I'm not sure where to go from here

Answer 1

I will assume those were exponents and you want:

f(x) = 3x^(1/3) + 6x^(4/3)
f ' (x) = x^(-2/3) + 8x^(1/3)
f '' (x) = (-2/3)x^(-5/3) + (8/3)x^(-2/3)

checking your work:
x^(-2/3) (1 + 8x) = 0
x = -1/8 , you had that!
but that is only the x of the extrema point
so f(-1/8) = 3(-1/8)^(1/3) + 6(-1/8)^(4/3)
= 3(-1/2) + 6(-1/16) = -15/8

so the extrema point is (-1/8, -15/8)
but right now we don't know if it is a max or a min

f ''(x) = 0 for points of inflection

(-2/3)x^(-5/3) + (8/3)x^(-2/3) = 0
x^(-5/3) - 4x^(-2/3) = 0
x^(-5/3)(1 - 4x) = 0
x = 1/4
You had -1/4

again , all you have is the x, we need the y
f(1/4) = .....
I will let you do that, not going to be "exact" since cube root of 4 is irrational

also f '' (-1/8) = positive, so (-1/8, -15/8) is a MINIMUM point.

You might want to check my arithmetic, I should have written it out on paper.