I need help with a curve dissection wkst. There are 12 problems. Some of them I need help with and others I just want to make sure I did them right. (AP Calculus)

Question

anonymous · Answer

We are instructed to find the critical points by using the first derrivative.
Then find the points of infliction using the second derrivative. 
Then to draw the curve on a graph

anonymous · Answer

the first problem is $$y=3x ^{2/3}-2x$$
the derrivative is (i think ) $$y \prime = 2x ^{1/3}-2$$
which means that the critical point is x=0?

the second derrivative is (once again, i am not positive) $$y \prime = \frac{ 2 }{ 3 }x ^{2/3}$$
makng the point of infliction x=0?

anonymous · Answer

Remember that $$ \frac{ d(x^n) }{ dx } = nx^{n-1}$$ you only decremented n by $$\frac{1}{3}$$

anonymous · Answer

so the derrivative would be $$y \prime = 2x ^{-1/3}-2$$

anonymous · Answer

exactly!

anonymous · Answer

how woudi find he critical pt for that?

anonymous · Answer

i know the derrivative is also equal to$$y \prime = 2(x ^{-1/3}-1)$$
if that helps any

anonymous · Answer

set y prime to 0 and solve as you would normally

anonymous · Answer

ok. so i get. . .$$0= 2x ^{-1/3}-2$$
add 2 to other side
$$2x ^{-1/3}=2$$
divide by 2
$$x ^{-1/3}=1$$
then i get stuck

anonymous · Answer

well$$x^{-1/3} = \frac{1}{x^{1/3}} = 1$$ right? 
try cubing both sides!

anonymous · Answer

did this help?