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).

I found the first and second derivative which i think is (8x+1)/(x^(2/3), and the second is 2(4x-1)/3x^(5/3). what now?

Assuming x is not zero, the fractions are zero when the numerators are zero.

So,

f'=0 when 8x+1 = 0, or x = -1/8
f"=0 when 4x-1 = 0, or x = 1/4

To find the x-coordinates of relative extrema and inflection points for the given function f(x) = 3x^(1/3) + 6x^(4/3), we need to analyze the first derivative (f'(x)) and the second derivative (f''(x)).

Let's start with the first derivative:

f'(x) = (8x + 1)/x^(2/3)

To find the critical points, we set f'(x) equal to zero and solve for x:

(8x + 1)/x^(2/3) = 0

Since the numerator is never zero, the critical point occurs when the numerator is equal to zero:

8x + 1 = 0
8x = -1
x = -1/8

So, one critical point is at x = -1/8.

Now, let's analyze the second derivative:

f''(x) = [2(4x - 1)] / (3x^(5/3))

To determine whether the critical point at x = -1/8 is a relative maximum, relative minimum, or an inflection point, we evaluate f''(x) at that point:

f''(-1/8) = [2(4(-1/8) - 1)] / (3(-1/8)^(5/3))
= [2((-1/2) - 1)] / (3(-1/8)^(5/3))
= [2((-3/2))] / (3(-1/8)^(5/3))
= (-3) / (3(-1/8)^(5/3))
= (-3) / (3(-1/2)^(5/3))
= (-3) / (3(-1/2)^(5/3))

Now, we need to consider the sign of f''(-1/8) to determine the nature of the critical point.

If f''(-1/8) > 0, it means the second derivative is positive, indicating a relative minimum.

If f''(-1/8) < 0, it means the second derivative is negative, indicating a relative maximum.

If f''(-1/8) = 0, it means the second derivative is zero, indicating a possible inflection point.

You can plug in the value and evaluate f''(-1/8) to determine the nature of the critical point at x = -1/8.

Once you have determined the nature of the critical point, you can continue analyzing the function to find other possible extrema and inflection points by finding the x-values where f'(x) = 0 or f''(x) = 0.