If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Integrated math 3

### Course: Integrated math 3>Unit 6

Lesson 3: Symmetry of functions

# Even and odd functions: Equations

When we are given the equation of a function f(x), we can check whether the function is even, odd, or neither by evaluating f(-x). If we get an expression that is equivalent to f(x), we have an even function; if we get an expression that is equivalent to -f(x), we have an odd function; and if neither happens, it is neither!

## Want to join the conversation?

• At , he says that the cube root of -1 is -1, but how? You can't find the root of a negative number, right?
• Maybe you confused cube root with square root. Although even roots of negative numbers cannot be solved with just real numbers, odd roots are possible. For example:
(-3)(-3)(-3)=cbrt(-27)
Even though you are multiplying a negative number, it is possible to obtain a negative answer because you are multiplying it with itself an odd number of times. Let's walk through it a little more slowly:
(-3)(-3)=9
9(-3)=-27
Similiarly, this will work for any number and for any odd exponent.
• hi, i don't understand why the function j(x) is neither odd or even. I thought
j(-x) was the opposite of j(x) which would make it odd. Can someone explain this to me? I am confusion
• it's because the 1 in the function is not negated.
• At , Khan says that 5/(3-(-x^4))=5/(3-x^4). I don't understand how it would not be 5/(3+x^4), as two negatives make a positive. Can somebody please explain to me how the negatives work in this question? Thanks in advance.
• It's not 5∕(3 − (−𝑥⁴)),
it's 5∕(3 − (−𝑥)⁴)

The exponent takes precedence over subtraction, so the first thing we do is simplify (−𝑥)⁴, which becomes 𝑥⁴.

Thus, 5∕(3 − (−𝑥)⁴) = 5∕(3 − 𝑥⁴)
• Why so complicated? Just plug a number into it.
• It's not always that easy. There is always a chance that substituting a number would yield the same result, despite them not actually being even functions. By solving algebraically, we can prove that they are certainly even or odd for every number.
• Yeah there is no cube root of -1, it will become an imaginary number, so the second equation is neither. Please tell me if I am right.
• -1 x -1 x -1 = -1
• Can you simplify x/(1-x)? What's the rule? What about (2^x) + (2^-x)?
• We can find equivalent functions using various algebraic rules, but whether or not they are simpler is subjective.

x/(1 - x)
= -1 * -x/(1 - x)
= -1* (-1 + 1 - x)/(1 - x)
= -1 * (-1/(1 - x) + (1 - x)/(1 - x))
= -1 * (-1/(1 - x) + 1)
= 1/(1 - x) - 1

2^x + 2^(-x)
= e^(ln(2^x)) + e^(ln(2^(-x)))
= e^(x * ln(2)) + e^(-x * ln(2))
= 2 * (e^(x * ln(2) + e^(-x * ln(2))/2)
= 2 * cosh(x * ln(2))
• why would you want to make f(x) to f(-x)?
• That appears to be the point of the video, this helps determine if a funciton is even, odd, or neither.
• Shouldn't we have checked whether each equation is BOTH odd AND even?
• The only function which is both odd and even is the zero function.
• My understanding is that an odd function must go through the origin. Is that correct? And if so, how can g(x) be an odd function when to take x = 0 would result in 1 / 0 which is undefined? Or is that ok as long as elsewhere in the function a valid answer is produce?