# Graphing a cubed root function

1. Nov 8, 2012

### frosty8688

1. The problem statement, all variables and given/known data

Sketch the curve of $\sqrt[3]{(x^{2} - 1)^{2}}$[/b]

2. Relevant equations

3. The attempt at a solution

I determined that the domain is all real numbers, the x-int. is ±1, the y-int. is 1, the function is symmetric about the y-axis, there are no asymptotes, and here is where I get lost. I took the first derivative of the function and found it to be equal to$\frac{2}{3}$ (x$^{2}$ - 1)$^{\frac{-1}{3}}$ * 2x = $\frac{4x}{3\sqrt[3]{x^{2}-1}}$ I know one of the critical numbers is 0 and I am wondering if the other numbers is ±1.

Last edited by a moderator: Nov 9, 2012
2. Nov 9, 2012

### ehild

What do you mean on "the x-int. is ±1, the y-int. is 1"?

Where is the function increasing and where is it decreasing? Try to sketch it. What happens at x=±1?

ehild

3. Nov 9, 2012

### frosty8688

At x=±1 the function lands on the x-axis. The function is decreasing on (-∞,0) and increasing on (0,∞).

4. Nov 9, 2012

### ehild

What happens in the interval (-1,1)?

ehild

5. Nov 9, 2012

### frosty8688

It concaves downward.

6. Nov 9, 2012

### SammyS

Staff Emeritus
Since the derivative is undefined at x = ±1, those also are critical numbers.

7. Nov 9, 2012

### ehild

What is the function at x=0?
Is it increasing or decreasing between -1 and 0, and between 0 and 1?

ehild

8. Nov 9, 2012

### frosty8688

It is increasing between -1 and 0 and decreasing between 0 and 1.

9. Nov 9, 2012

### frosty8688

At x=0 the value is 1.

10. Nov 9, 2012

### ehild

Well, so it is decreasing in (-∞,-1), increasing between -1 and 0, decreasing between 0 and 1, increasing in (1,∞). The function is 1 at x=0 and 0 at ±1. Can you sketch it?

ehild

11. Nov 9, 2012

### frosty8688

I have another question. I am trying to find the second derivative of the function and this is what I have. $f"(x) = \frac{4*3\sqrt[3]{x^{2}- 1}- 4x*3*\frac{1}{3\sqrt[3]{x^{2}-1}}}{9\sqrt[3]{(x^{2}-1)}^{2}} = \frac{12\sqrt[3]{x^{2}-1}- 12x}{27\sqrt[3]{(x^{2}-1)^{2}}\sqrt[3]{x^{2}-1}} = \frac{12\sqrt[3]{x^{2}-1}- 12x}{27(x^{2}-1)}$. I am wondering if this is right so far.

12. Nov 9, 2012

### SammyS

Staff Emeritus
Incorrect.

The derivative of $\ \ 3\sqrt[3]{x^2-1}\ \$ is $\displaystyle \ \ \frac{2x}{\sqrt[3]{(x^2-1)^2}}\ .$

13. Nov 9, 2012

### frosty8688

I'm talking about the second derivative.

14. Nov 9, 2012

### SammyS

Staff Emeritus
So was I.

In taking the derivative, using the quotient rule, you took the derivative of the denominator of the first derivative incorrectly.

15. Nov 9, 2012

### frosty8688

So the equation becomes $f"(x) = \frac{4*3\sqrt[3]{x^{2}-1}- 4x*\frac{2x}{\sqrt[3]{(x^{2}-1)^{2}}}}{9\sqrt[3]{(x^{2}-1)^{2}}} = \frac{12\sqrt[3]{x^{2}-1}- 8x^{2}}{9(x^{2}-1)\sqrt[3]{x^{2}-1}} = \frac{12-8x^{2}}{9(x^{2}-1)}$

16. Nov 10, 2012

### SammyS

Staff Emeritus
Not quite.

The first step is correct.

17. Nov 10, 2012

### frosty8688

So what did I do wrong.

18. Nov 10, 2012

### SammyS

Staff Emeritus
$\displaystyle f"(x) = \frac{4*3\sqrt[3]{x^{2}-1}- 4x*\frac{2x}{\sqrt[3]{(x^{2}-1)^{2}}}}{9\sqrt[3]{(x^{2}-1)^{2}}} = \frac{12\sqrt[3]{(x^{2}-1)^3}- 8x^{2}}{9(x^{2}-1)\sqrt[3]{x^{2}-1}}= \ \ \dots$

19. Nov 10, 2012

### frosty8688

Is there a problem between the second and third steps.

20. Nov 10, 2012

### frosty8688

So the cube root on the top would disappear and it would be $\frac{4x^{2}-1}{9(x^{2}-1)\sqrt[3]{x^{2}-1}}$

Last edited: Nov 10, 2012