Graphing a cubed root function

[frosty8688]

Homework Statement

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

Homework Equations

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.

 

[ehild]

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

Homework Equations

3. 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

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

 

[frosty8688]

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

 

[ehild]

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

ehild

 

[frosty8688]

It concaves downward.

 

[SammyS]

I know one of the critical numbers is 0 and I am wondering if the other numbers is ±1 [/b]

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

 

[ehild]

It concaves downward.

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

ehild

 

[frosty8688]

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

 

[frosty8688]

At x=0 the value is 1.

 

[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

 

[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.

 

[SammyS]

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.

Incorrect.

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

 

[frosty8688]

I'm talking about the second derivative.

 

[SammyS]

I'm talking about the second derivative.

So was I.

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

 

[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)}

 

[SammyS]

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)}

Not quite.

The first step is correct.

 

[frosty8688]

So what did I do wrong.

 

[SammyS]

\displaystyle f&quot;(x) <br /> = \frac{4*3\sqrt[3]{x^{2}-1}- 4x*\frac{2x}{\sqrt[3]{(x^{2}-1)^{2}}}}{9\sqrt[3]{(x^{2}-1)^{2}}} <br /> = \frac{12\sqrt[3]{(x^{2}-1)^3}- 8x^{2}}{9(x^{2}-1)\sqrt[3]{x^{2}-1}}= \ \ \dots<br />

 

[frosty8688]

Is there a problem between the second and third steps.

 

[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}}

 

[SammyS]

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}}

When the cube root in the numerator disappears, the numerator becomes 12(x^2-1)-8x^2\ .


You simplified that incorrectly.

 

[frosty8688]

I forgot to distribute the 12.

 

[frosty8688]

So the equation \frac{4(x^{2}-3)}{9(x^{2}-1)\sqrt[3]{x^{2}-1}} has critical numbers at ±\sqrt{3} or ±1.

 

[HallsofIvy]

\frac{ab+ c}{b}\ne a+ c

 

[SammyS]

So the equation \frac{4(x^{2}-3)}{9(x^{2}-1)\sqrt[3]{x^{2}-1}} has critical numbers at ±\sqrt{3} or ±1.

It's not clear what equation you're referring to.

The first derivative is undefined at x = ±1 . Of course this means that the second derivative is also undefined at x = ±1 .

The second derivative is zero at x=\pm\sqrt{3}, so that the function has inflection points there.

 

[frosty8688]

That's what I figured.

 

[frosty8688]

That is also where it concaves down. Right?

 

[SammyS]

That is also where it concaves down. Right?

That's where it changes concavity, from up to down, or vice versa .