# Thread: Precal - Even and Odd Functions

1. ## Precal - Even and Odd Functions

Need help. And if you could, explain in detail what to do. Here it is:

Determine algebraically whether each of the following functions is even, odd, or neither.

3 = Cubed Root
\| = Square Root

f (x) = 3\|x

I guess it would be hard to read but it's F of X equals the cubed root of X. How would I solve this equation?

2. Originally Posted by Monkiwrench
Need help. And if you could, explain in detail what to do. Here it is:

Determine algebraically whether each of the following functions is even, odd, or neither.

3 = Cubed Root
\| = Square Root

f (x) = 3\|x

I guess it would be hard to read but it's F of X equals the cubed root of X. How would I solve this equation?
a function $f(x)$ is even if $f(-x) = f(x)$

a function $f(x)$ is odd if $f(-x) = -f(x)$

you could have typed f(x) = cuberoot(x) or f(x) = x^(1/3)

3. Right, Jhevon. I know that much. But what I need help on doing (step by step) is 'testing' or putting it through the process in which we take f(x)=Cubed Root of X and see if it fits the Even, Odd, or Neither profile. Here's one I already know how to do:

f(x) = x / x(squared) - 1

f(-x) = -x / -x(squared) -1 = -x / x(squared) - 1

-f(x) = - x / x(squared) - 1 = - x / x(squared) - 1

And, no, I don't know how to use the /math symbols.

But no, it's not how to get started, it's figuring this certain problem out that's stumping me.

4. Originally Posted by Monkiwrench
Right, Jhevon. I know that much. But what I need help on doing (step by step) is 'testing' or putting it through the process in which we take f(x)=Cubed Root of X and see if it fits the Even, Odd, or Neither profile. Here's one I already know how to do:

$f(x) = x / x(squared) - 1

f(-x) = -x / -x(squared) -1 = -x / x(squared) - 1

-f(x) = - x / x(squared) - 1 = - x / x(squared) - 1$

And, no, I don't know how to use the /math symbols.

But no, it's not how to get started, it's figuring this certain problem out that's stumping me.
you can't just make up commands and expect them to work. to write x squared you must type x^2, to write a fraction you must use \frac {}{} where the numerator goes in the first {} and the denominator in the second. see our LaTex tutorial here

the formulas i gave you tells you the steps. just replace $x$ with $-x$ and simplify. if you get back the original function, then the function is even, if you get the negative of the function, then it is odd. otherwise, it is neither.

example: $f(x) = x^2$ is even, since $f(-x) = (-x)^2 = x^2 = f(x)$

example 2: $f(x) = x^3$ is odd, since $f(-x) = (-x)^3 = -x^3 = -f(x)$

what is confusing you here? it's the same procedure

5. Good stuff. I'm understanding now how it's done, but with my problem, it's not X squared or cubed, it's $\sqrt{x}$. And I don't know how to work that out.

6. Originally Posted by Monkiwrench
Good stuff. I'm understanding now how it's done, but with my problem, it's not X squared or cubed, it's $\sqrt{x}$. And I don't know how to work that out.
if $f(x) = \sqrt {x}$, for $x > 0$, then $f(-x) = \sqrt {-x} = \sqrt {x}~i \ne f(x) \mbox { or} -f(x)$, therefore, it is neither even nor odd.

however, $\sqrt [3] {-x}$ is not imaginary. what do you think it is? what number can we multiply by itself 3 times and get $-x$?

7. what number can we multiply by itself 3 times and get ?
That's... not possible is it? Multiplying a number 3 times to get a variable?

8. Originally Posted by Monkiwrench
That's... not possible is it? Multiplying a number 3 times to get a variable?
haha, well, variables represent numbers don't they? it's just that the numbers can change so we replace them with letters. how do you think we get $x^2$? we multiply two numbers, $x$ and $x$, $x \cdot x = x^2$. what number is $x$? i don't know, it could be any number, it depends on what i'm using it for

9. $f(x) = \sqrt[3]{x}$ is the problem. What you're using x for is to see if the equation falls under the even, odd or neither category.

But multiplying $\frac{x}{3}$ 3 times would give you X, right?

10. Originally Posted by Monkiwrench
$f(x) = \sqrt[3]{x}$ is the problem. What you're using x for is to see if the equation falls under the even, odd or neither category.

But multiplying $\frac{x}{3}$ 3 times would give you X, right?
that is not multiplying a number by itself 3 times. x/3 and 3 are TWO DIFFERENT numbers

$\sqrt [3] {x} = x^{1/3}$

since $\left( x^{1/3} \right)^3 = x^{1/3} \cdot x^{1/3} \cdot x^{1/3} = x^{1/3 + 1/3 + 1/3} = x$

11. I'm gonna try and figure this one out on my own. Correct me in the spots necessary...

Gonna test it out and see if it's even...

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

It's not the same as the equation, therefore it couldn't be even, right?

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

That's the same as the original equation, therefore it couldn't be odd, right?

So it's neither?

12. Originally Posted by Monkiwrench

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

That's the same as the original equation, therefore it couldn't be odd, right?

So it's neither?
what are you talking about? you just showed that $f(-x) = -f(x)$, by definition, what does that mean? well, not really, you should have showed that

i'll just tell you. $f(-x) = \sqrt [3] {-x} = -\sqrt [3] {x} = -f(x)$ since $\left( -x^{1/3} \right)^3 = \left(- x^{1/3} \right) \cdot \left(- x^{1/3} \right) \cdot \left(- x^{1/3} \right) = -x^{1/3 + 1/3 + 1/3} = -x$

13. Making the equation odd, right? Origin symmetrical?

14. Originally Posted by Monkiwrench
Making the equation odd, right? Origin symmetrical?
yes, that means it's odd, which is the same as being symmetric about the origin (i'm pretty sure ....if you don't have faith in me you can look it up. a quick google search can put all doubts to rest)

15. You're awesome, Jhevon.

Okay, I have another thing to ask. I want to see how YOU would, step by step, figure out if the following equation would be classified as an even or odd function or neither.

$
f(x) = x^2 - x
$

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