Find relative extrema?

Question

f(x)= 2x+3x^(2/3)

Answer 1

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

solve for 2 + 2 x^(-1/3)=0

-1 = x
0 = x

Get f'' if you're interested in finding out if they're rel. max or min.

Answer 2

You have a relative max where f'(x) = 0:
f'(x) = 2 + 2x^(-1/3) = 0
2x^(-1/3) = -2
x^(-1/3) = -1
x = -1

You know it's a max (aside from looking at the graph), since f"(x) = (-2/3)x^(-4/3) and f"(-1) = -2/3 < 0.

You also have a min at x=0, but it's a sharp point, where you have a vertical tangent, since f'(x)→∞ as x→0.

Answer 3

Compute the first derivative function and equate it to 0. Algebraically isolate x.

0 = f'(x) = 2 + 3(2/3)x^(2/3 - 3/3).

0 = 2 + 2x^(-1/3).

0 = 2 + 2/cube root(x).

-2 = 2/cube root(x).

-2/2 = 1/cube root(x).

-1 = 1/cube root(x).

-1 = cube root(x).

(-1)^3 = x.

-1 = x.



Evaluate the original equation with x = -1

f(x = -1) = 2(-1) + 3(-1)^(2/3)

f(1) = -2 + (-3)^(2/3)

= -2 - cube root[(-3)^2)]

= -2 - cube root(9)

= an extremum.