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.

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- Sep 13th 2009, 07:54 AMmwokWhat 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. - Sep 13th 2009, 08:11 AMskeeter
the function you cite is actually $\displaystyle f(x) = -\sqrt[3]{x}$

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

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

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

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

$\displaystyle f(x) = -\sqrt[3]{x}$ is an odd function symmetric to the origin. - Sep 13th 2009, 08:19 AMmwok
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. - Sep 13th 2009, 08:26 AMskeeter
$\displaystyle f(x)$ is odd if $\displaystyle f(-x) = -f(x)$

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

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

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

free graph program ...

Graph - Sep 13th 2009, 08:30 AMPlato
The

__names__*odd & even*most likely came from exponents.

But exponents really do not determine much.

For example: $\displaystyle \cos(x)$ is even and $\displaystyle \sin(x)$ is odd.

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

If $\displaystyle -g(x)=g(-x)$ then $\displaystyle g$ is odd. - Sep 13th 2009, 08:35 AMmwok
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.

- Sep 13th 2009, 08:54 AMPlato
- Sep 13th 2009, 10:36 AMmwok
So, the answer is simply that the symmetry is about the origin?

- Sep 13th 2009, 10:40 AMPlato