Symmetry of cubed root of x^2 = 1

[katie9426]

I need to figure out if the graph of the cubed root of x^2=1 is symmetric about the x-axis, y-axis, or origin, and sadly enough I don't remember how to do that. Thanks for your help!

 

[skeeter]

forget what you learned in precalculus?

if f(-x) = f(x), then f(x) is an even function and is symmetric to the y-axis.

if f(-x) = -f(x), then f(x) is an odd function and is symmetric to the origin.

a graph that is symmetric to the x-axis is not a function ... if so, then y = f(x) and -y = f(x) ... each x-value maps to two opposite y-values.

 

[katie9426]

Yes and I'm still confused because wouldn't that function be none of those?

 

[stapel]

...the cubed root of x^2=1...

I've heard of a value cubed, or the cube root, but not "the cubed root". Also, graphs usually have an "x" and a "y"; what you have posted is not a function, but is a one-variable equation.

Do you mean any of the following?

. . . . .f(x) = (sqrt[x² - 1])³

. . . . .f(x) = (sqrt[x² + 1])³

. . . . .f(x) = (sqrt[x²]³ - 1

. . . . .f(x) = (sqrt[x²]³ + 1

. . . . .f(x) = cbrt[x²] - 1

. . . . .f(x) = cbrt[x²] + 1

. . . . .f(x) = cbrt[x² - 1]

. . . . .f(x) = cbrt[x² + 1]

Or something else?

When you reply, please show all of your work and reasoning. Thank you.

Eliz.

 

[katie9426]

f(x) = cbrt[x2 + 1] is what I meant. Sorry I didn't realize the typo. I typed the function into y= on my calculator, and it appears to be symmetric across the y axis. I was just wondering if that was correct or if there was a better way to try and find the symmetry. I didn't really understand the whole putting in negative/positve x's or y's because unless I was doing it wrong I always got a different function, which was what led me to believe it wasnt symetric about anything.

 

[stapel]

I typed the function into y= on my calculator.... I was just wondering...if there was a better way to try and find the symmetry.

Yes: Use the definition and algorithm, as provided by the tutor.

You say you got a result of "neither": Please reply showing your work and reasoning. Thank you.

Eliz.

 

[katie9426]

I got that it was symmetric about the y only. Is this correct?

 

[soroban]

Hello, Katie!

I got that it was symmetric about the y-axis only.

Right!


Here's that ± rule in simple terms.

Replace \(\displaystyle x\) with -\(\displaystyle x\).
If we get the original function, the graph is symmetric to the y-axis.

We have: \(\displaystyle \:f(x)\:=\:\sqrt[3]{x^2\,+\,1}\)

Then: \(\displaystyle \:f(-x) \:=\:\sqrt[3]{(-x)^2\,+\,1} \:=\:\sqrt[3]{x^2\,+\,1}\;\;\)← .the original function

Therefore, \(\displaystyle f(x)\) is symmetric to the y-axis.


Here's an "eyeball" version.

If all the exponents on the \(\displaystyle x\)'s are even, it's symmetric to y-axis.