# Math Help - What type of function is this?

1. ## What type of function is this?

y = -x^(1/3)

What type of function is this? Simplified it is y = i*cube root(x)

I'm trying to find the symmetry using the formula x = -b/2a.

2. Originally Posted by mwok
y = -x^(1/3)

What type of function is this? Simplified it is y = i*cube root(x)

I'm trying to find the symmetry using the formula x = -b/2a.
the function you cite is actually $f(x) = -\sqrt[3]{x}$

did you mean $y = (-x)^{\frac{1}{3}}$ ?

if so, this is the same as $f(x) = -\sqrt[3]{x}$

note that $(-x)^{\frac{1}{3}} = (-1)^{\frac{1}{3}} \cdot x^{\frac{1}{3}} = -1 \cdot x^{\frac{1}{3}}$

no i is involved ... $(-1)^{\frac{1}{2}} = i$

$f(x) = -\sqrt[3]{x}$ is an odd function symmetric to the origin.

3. But how did you find the symmetry and odd/even?

I know that an even/odd power corresponds to a even/odd function, however...the power in this case is a fraction. Without graphing, how can I determined this?

BTW, what graphing tool did you use to generate that graph?

Thanks.

4. Originally Posted by mwok
But how did you find the symmetry and odd/even?

I know that an even/odd power corresponds to a even/odd function, however...the power in this case is a fraction. Without graphing, how can I determined this?

BTW, what graphing tool did you use to generate that graph?

Thanks.
$f(x)$ is odd if $f(-x) = -f(x)$

$f(x) = \sqrt[3]{-x} = -\sqrt[3]{x}$

$f(-x) = \sqrt[3]{-(-x)} = \sqrt[3]{x}$

$f(-x) = \sqrt[3]{x}$ is the opposite of $f(x) = -\sqrt[3]{x}$ , therefore $f(x)$ is odd.

free graph program ...

Graph

5. Originally Posted by mwok
But how did you find the symmetry and odd/even?
I know that an even/odd power corresponds to a even/odd function, however...the power in this case is a fraction. Without graphing, how can I determined this?
The names odd & even most likely came from exponents.
But exponents really do not determine much.
For example: $\cos(x)$ is even and $\sin(x)$ is odd.

Learn the actual definitions: If $f(x)=f(-x)$ then $f$ is even.
If $-g(x)=g(-x)$ then $g$ is odd.

6. Thanks, how do you find the symmetry? I cannot use x = -b/2a because that is for polynomials and the function above is not a polynomial.

7. Originally Posted by mwok
Thanks, how do you find the symmetry?
First do as I suggested: Learn the basic definitions.
Then it is simple: odd functions are symmetric about the origin; even functions are symmetric about the y-axis.

8. So, the answer is simply that the symmetry is about the origin?

9. Originally Posted by mwok
So, the answer is simply that the symmetry is about the origin?
Well of course, if it is an odd function.
$f(x) = - x^{\frac{1}
{3}} \; \Rightarrow \;f( - x) = - \left( { - x} \right)^{\frac{1}
{3}} = x^{\frac{1}
{3}} = - f(x)$
so it is odd.