- #1

aeromat

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## Homework Statement

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

## 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

[tex]0 = (x)^{1/3}*(x+3)^{2/3}[/tex]

[tex]0 = (x+3)^{2/3}[/tex]

[tex]0^3 = ((x+3)^{2/3})^3[/tex]

[tex]\sqrt{0} = \sqrt{(x+3)^2}[/tex]

[tex]0 = (x + 3)[/tex]

[tex]-3 = x[/tex]

Therefore, x-intercept at x = -3.

[tex]y = (0)^{1/3}*(3)^{2/3}[/tex]

[tex]y = 0[/tex]

Therefore, the graph crosses the origin; (0,0)

First Derivative:

[tex]g'(x) = a'(b) + a(b)'[/tex]

[tex]a = (x)^{1/3}[/tex]

[tex]a' = \frac{1}{3}*(x)^{-2/3}[/tex]

[tex]b = (x+3)^{2/3}[/tex]

[tex]b' = \frac{2}{3}*(x+3)^{-1/3}[/tex]

The extreme value retrieved from the first derivative:

[tex]0 = (x+1)[/tex]

[tex]-1 = x [/tex]Simplifying for the first derivative, and I get:

[tex]g'(x) = \frac{x+1}{x^{2/3}*(x+3)^{1/3}}[/tex]

*?: 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...