# Thread: Functions: Even, Odd or Neither

1. ## Functions: Even, Odd or Neither

Determine algebraically if each function is even, odd, or neither.

1. F(x) = cuberoot{x}

2. F(x) = 2x/|x|

3. f(x) = x + |x|

Like always, seeking the needed steps.

2. ## Re: Functions: Even, Odd or Neither

A function is even if:

$\displaystyle f(-x)=f(x)$

A function is odd if:

$\displaystyle f(-x)=-f(x)$

Using these rules, what do you conclude about the given functions?

3. ## Re: Functions: Even, Odd or Neither

An example:
\begin{align}\text{let} \quad f(x) &= x^2 \\ \text{then} \quad f(-x) &= (-x)^2 \\ &= \big((-1)x\big)^2 \\ &= (-1)^2 x^2 \\ &= (1)x^2 \\ &= x^2 \\ &= f(x) \end{align}
So $f(x)$ Is even.

4. ## Re: Functions: Even, Odd or Neither

can we at least assume that you know that |x|= x if $\displaystyle x\ge 0$ and |x|= -x if x< 0.

From that
2) $\displaystyle f(x)= \frac{2x}{|x|}$
if $\displaystyle x\ge 0$ $\displaystyle f(x)= \frac{2x}{x}= 2$ and if $\displaystyle x< 0$, $\displaystyle f(x)= \frac{2x}{-x}= -2$.

3) $\displaystyle f(x)= x+ |x|$
if $\displaystyle x\ge 0$ $\displaystyle f(x)= x+ x= 2x$ and if $\displaystyle x< 0$, $\displaystyle f(x)= x+ (-x)= 0$.

5. ## Re: Functions: Even, Odd or Neither

Originally Posted by harpazo
Determine algebraically if each function is even, odd, or neither.
2. $F(x) = \dfrac{2x}{|x|}$
Like always, seeking the needed steps.
\displaystyle \begin{align*}F(-x) &= \dfrac{-2x}{|-x|} \\&= -\dfrac{2x}{|x|} \\&= -F(x)\end{align*}

6. ## Re: Functions: Even, Odd or Neither

Originally Posted by MarkFL
A function is even if:

$\displaystyle f(-x)=f(x)$

A function is odd if:

$\displaystyle f(-x)=-f(x)$

Using these rules, what do you conclude about the given functions?
1. F(x) = cuberoot{x}

F(-x) = cuberoot{-x}

Result: neither

2. F(x) = 2x/|x|

F(-x) = 2(-x)/|-x| = -2x/x = -2

Result: neither

3. f(x) = x + |x|

f(-x) = -x + |-x| = -x + x = 0

Result: neither

7. ## Re: Functions: Even, Odd or Neither

Originally Posted by harpazo
1. F(x) = cuberoot{x}

F(-x) = cuberoot{-x}

Result: neither

2. F(x) = 2x/|x|

F(-x) = 2(-x)/|-x| = -2x/x = -2

Result: neither

3. f(x) = x + |x|

f(-x) = -x + |-x| = -x + x = 0

Result: neither
Careful! |-x| is not equal, in general, to x.

$\displaystyle |x| = \begin{cases} -x & x < 0 \\ x & 0 < x \end{cases}$

We don't know if x is negative or positive. So the only thing you can say is
$\displaystyle F(-x) = \dfrac{2(-x)}{|-x|} = \dfrac{-2x}{|x|} = -\dfrac{2x}{|x|} = -F(x)$

As HallsofIvy and Plato have already told you.

You have a similar problem with #3.

-Dan

8. ## Re: Functions: Even, Odd or Neither

Originally Posted by harpazo
1. F(x) = cuberoot{x}

F(-x) = cuberoot{-x}

Result: neither
Actually, given:

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

We have:

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

So, this function is odd.

9. ## Re: Functions: Even, Odd or Neither

Originally Posted by MarkFL
Actually, given:

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

We have:

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

So, this function is odd.
I did not know that negative 1 can be pulled out of the radicand.

10. ## Re: Functions: Even, Odd or Neither

Originally Posted by topsquark
Careful! |-x| is not equal, in general, to x.

$\displaystyle |x| = \begin{cases} -x & x < 0 \\ x & 0 < x \end{cases}$

We don't know if x is negative or positive. So the only thing you can say is
$\displaystyle F(-x) = \dfrac{2(-x)}{|-x|} = \dfrac{-2x}{|x|} = -\dfrac{2x}{|x|} = -F(x)$

As HallsofIvy and Plato have already told you.

You have a similar problem with #3.

-Dan
So, the answer to question 2 is odd.

11. ## Re: Functions: Even, Odd or Neither

Question 3

f(x) = x + |x|

f(-x) = -x + |-x|

f(-x) = -x + |x|

The function is odd.

12. ## Re: Functions: Even, Odd or Neither

Originally Posted by harpazo
Question 3
$f(x) = x + |x|$
The function is odd.
$-f(x) = -(x + |x|)=-x-|x|\ne -x+|-x|=f(-x)$ So is it odd?

13. ## Re: Functions: Even, Odd or Neither

Originally Posted by harpazo
I did not know that negative 1 can be pulled out of the radicand.
\begin{align} y=\sqrt[3]{x} \implies y^3 &= x \\
\text{so}\;{-x} &= -y^3 = (-1)y^3 \\ &= (-1)^3y^3 = \big((-1)y\big)^3 = (-y)^3 \\
\text{thus}\; \sqrt[3]{-x} &= -y = -\sqrt[3]{x} \end{align}

14. ## Re: Functions: Even, Odd or Neither

Originally Posted by harpazo
I did not know that negative 1 can be pulled out of the radicand.
Note: $\displaystyle \sqrt[n]{x} = (x)^{1/n}$.

$\displaystyle \sqrt[n]{-x} = - \sqrt[n]{x}$ only if n is an odd number.

-Dan

15. ## Re: Functions: Even, Odd or Neither

Originally Posted by Plato
$-f(x) = -(x + |x|)=-x-|x|\ne -x+|-x|=f(-x)$ So is it odd?
It's neither.

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