Sketch this curve (two multipled root functions)

[aeromat]

Homework Statement

Sketch the curve defined by g(x) = x^{1/3}*(x+3)^{2/3}

Homework Equations

First Derivative
Second Derivative

The Attempt at a Solution

I attempted to use the algorithm for curve sketching:
1) Find the intercepts
2) First Derivative; look for extrema and points where f'(x) DNE
3) Second Derivative; look for each region's concavity according to point of inflection found, and where f''(x) DNE
4) Sketch the curve

First, I made the function y = 0
0 = (x)^{1/3}*(x+3)^{2/3}
0 = (x+3)^{2/3}
0^3 = ((x+3)^{2/3})^3
\sqrt{0} = \sqrt{(x+3)^2}
0 = (x + 3)
-3 = x
Therefore, x-intercept at x = -3.

y = (0)^{1/3}*(3)^{2/3}
y = 0
Therefore, the graph crosses the origin; (0,0)

First Derivative:
g'(x) = a'(b) + a(b)'
a = (x)^{1/3}
a' = \frac{1}{3}*(x)^{-2/3}
b = (x+3)^{2/3}
b' = \frac{2}{3}*(x+3)^{-1/3}

The extreme value retrieved from the first derivative:
0 = (x+1)
-1 = xSimplifying for the first derivative, and I get:
g'(x) = \frac{x+1}{x^{2/3}*(x+3)^{1/3}}

?: What does it mean when you have a divide by zero error in the first derivative?

The second derivative is much more complicated. I finished it but I first need to understand as to what do I have to do when I get it in the first place...

 

[tiny-tim]

hi aeromat!

?: What does it mean when you have a divide by zero error in the first derivative?

it means the "slope" is "vertical"