Question

Find all the points of inflection of the function.
y=x^1/3(x-4)

Answer 1

So your goal is to find the second derivative.
I will give you a new goal when you have done that.

Answer 2

the second derivative is what i feel I keep getting wrong because for the first I get \[x^{1/3}+\frac{ 1 }{ 3 } x^2 -4\] and for the second i get \[\frac{ 2 }{ 3 }x ^{4/3}+\frac{ 1 }{ 9 }x^2-\frac{ 4 }{ 3 }\]

Answer 3

I feel like thats wrong

Answer 4

Is your y this:
\[y=x^\frac{1}{3}(x-4)?\]

Answer 5

no its \[x ^{1/3}(x-4)\]...the 1/3 is the exponent

Answer 6

Ok so we used the product rule:
\[y'=x^\frac{1}{3} \cdot (1-0)+\frac{1}{3}x^{\frac{1}{3}-1} \cdot (x-4)\]
Is this what you did for the first derivative?

Answer 7

\[y'=x^\frac{1}{3}(1)+\frac{1}{3}x^\frac{-2}{3}(x-4)\]
\[y'=x^\frac{1}{3}+\frac{1}{3}x^\frac{1}{3}-\frac{4}{3}x^\frac{-2}{3}\]
?

Answer 8

yea thats what I got the first time I tried the problem

Answer 9

I don't you did. Unless you typed the wrong thing.

Answer 10

Anyways we do have like terms in y'. Combine those.

Answer 11

no, what I'm trying to say is that the answer you posted earlier was the same as what I got when I redid the first derivative

Answer 12

so Now i just need to find the second derivative

Answer 13

yep so you are trying to find the derivative of this:
\[y'=\frac{4}{3}x^\frac{1}{3}-\frac{4}{3}x^\frac{-2}{3}\]

Answer 14

ok so then I get \[\frac{ 4 }{ 9 }x ^{-2/3}+ \frac{ 8 }{ 9 }x^{-5/3}\].....right

Answer 15

right

Answer 16

Objective write as one fraction
Main objective to find when y'' is 0 and y'' dne.
\[y''=\frac{4}{9x^\frac{2}{3}}+\frac{8}{9x^\frac{5}{3}}\]
Combine fractions:
\[y''=\frac{4 x^\frac{3}{3}}{9x^\frac{2}{3} x^\frac{2}{3}}+\frac{8}{9x^\frac{5}{3}}\]

Answer 17

\[y''=\frac{4x+8}{9x^\frac{5}{3}}\]

Answer 18

ok so now i set it equal to zero right

Answer 19

thanks

Answer 20

Main objective to find when y'' is 0 and y'' dne.

Then you see if the concavity switchs at both of those numbers.