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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...