Finding the value of x in y' and y"

[SeekerOfDragons]

I am trying to find the local maxima/minima, inflection points, and direction of concavity (up/down)

My main issue is finding the inflection points because I'm having difficulty solving for x in the y" equation.

I am having trouble finding the y" of the following equation:
y=3/4(x^2-1)^2/3

I managed to get y' as:
y' = (3/4)(2/3)(x^2-1)^-1/3(2x) = x(x^2-1)^-1/3 OR x/(x^2-1)^1/3
x = -1, 0, 1

I have tried both the Product Rule and Quotient rule to differentiate y' to y" but have hit a wall.
using the Product rule I've gotten this far:

y" = x[-1/3(x^2-1)^-4/3(2x)] + 1(x^2-1)^-1/3

any assistance in reducing y" and solving for x will be much appreciated. once I have x, i can finish the problem without any issues. I just can't figure out how to reach the answer given in the back of my book.

r/

SoD

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ \frac{3(x^{2}-1)^{2/3}}{4}\)

\(\displaystyle f \ ' \ (x) \ = \ \frac{x}{(x^{2}-1)^{1/3}}, \ rel. \ Max \ at \ (0,3/4), \ rel. \ Min \ at \ (\pm1,0)\)

\(\displaystyle f \ " \ (x) \ = \ \frac{x^{2}-3}{3(x^{2}-1)^{4/3}}, \ Pts. \ of \ Inf. \ \bigg(\pm\sqrt3,\frac{3(2)^{2/3}}{4}\bigg)\)

\(\displaystyle Concave \ up: \ (-\infty,-\sqrt3]U[\sqrt3,\infty), \ Concave \ down: \ [-\sqrt3,\sqrt3]\)

\(\displaystyle See \ graph.\)

[attachment=0:umoulj7m]def.jpg[/attachment:umoulj7m]

 

[SeekerOfDragons]

How did you get to the final version of y"? That's where I'm having issues. getting y' converted to y". any chance of showing the steps to get from y' to y"?

 

[BigGlenntheHeavy]

\(\displaystyle f \ ' \ (x) \ = \ \frac{x}{(x^{2}-1)^{1/3}}\)

\(\displaystyle f \ " \ (x) \ = \ \frac{(x^{2}-1)^{1/3}(1)-x(1/3)(x^{2}-1)^{-2/3}(2x)}{(x^{2}-1)^{2/3}}\)

\(\displaystyle = \ \frac{(x^{2}-1)^{1/3}-\frac{2x^{2}}{3(x^{2}-1)^{2/3}}}{(x^{2}-1)^{2/3}}\)

\(\displaystyle = \ \frac{3(x^{2}-1)-2x^{2}}{3(x^{2}-1)^{4/3}} \ = \ \frac{x^{2}-3}{3(x^{2}-1)^{4/3}}\)

 

[SeekerOfDragons]

Thanks. appreciate it. was close a few times, just forgot to multiply the 3 through and then subtract the x^2. makes sense now.