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Integrated math 3
Course: Integrated math 3 > Unit 6
Lesson 3: Symmetry of functions- Function symmetry introduction
- Function symmetry introduction
- Even and odd functions: Graphs
- Even and odd functions: Tables
- Even and odd functions: Graphs and tables
- Even and odd functions: Equations
- Even and odd functions: Find the mistake
- Even & odd functions: Equations
- Symmetry of polynomials
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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!
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At2:16, he says that the cube root of -1 is -1, but how? You can't find the root of a negative number, right?(13 votes)
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.(30 votes)
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(12 votes)
it's because the 1 in the function is not negated.(3 votes)
At1:13, 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.(4 votes)
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)(9 votes)
Why so complicated? Just plug a number into it.(4 votes)- 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.(6 votes)
- 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.(3 votes)
- -1 x -1 x -1 = -1(7 votes)
Can you simplify x/(1-x)? What's the rule? What about (2^x) + (2^-x)?(3 votes)
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))(6 votes)
- why would you want to make f(x) to f(-x)?(3 votes)
That appears to be the point of the video, this helps determine if a funciton is even, odd, or neither.(4 votes)
Shouldn't we have checked whether each equation is BOTH odd AND even?(2 votes)- The only function which is both odd and even is the zero function.(4 votes)
- 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?(5 votes)
- I cant help you(0 votes)
f(x) = x/e^x -1 + 2/x + cos x.
Is this function even or odd?(2 votes)
This function seems like a whole bunch of different functions mashed together, so there's a good chance it will be neither even nor odd (A function is even if f(-x) = f(x), even functions are the same when reflected across the y-axis. A function is odd when f(-x) = -f(x); odd functions look the same when rotated 180 degrees). We can test if a function is even or odd by plugging in (-x) for x and seeing what happens:
f(-x) = (-x / (e^(-x) - 1) + 2/(-x) + cos(-x)
At least to me, it doesn't look like you can simplify it further in any way besides taking out the negative in -x / -1 + e^(-x). So this function is neither even nor odd.(4 votes)
2:16
Video transcript
- [Instructor] We are asked are the following functions
even, odd, or neither? So pause this video and try
to work that out on your own before we work through it together. All right, now let's just remind ourselves of a definition for
even and odd functions. One way to think about it is what happens when you take f of negative x? If f of negative x is equal
to the function again, then we're dealing with an even function. If we evaluate f of negative x, instead of getting the function, we get the negative of the function, then we're dealing with an odd function. And if neither of these
are true it is neither. So let's go to this first
one right over here, f of x is equal to five over
three minus x to the fourth, and the best way I can
think about tackling this is let's just evaluate what f of negative x would be equal to. That would be equal to
five over three minus and everywhere we see an x, we're gonna replace
that with a negative x, to the fourth power. Now what is negative
x to the fourth power? Well if you multiply a
negative times a negative times a negative, how
many times did I do that? If you take a negative
to the fourth power, you're going to get a positive, so that's going to be
equal to five over three minus x to the fourth, which
is once again equal to f of x and so this first one right over here, f of negative x is equal to
f of x, it is clearly even. Let's do another example. So this one right over here, g of x, let's just evaluate g of negative x and at any point, you feel inspired and you didn't figure
it out the first time, pause the video again and try
to work it out on your own. Well g of negative x is
equal to one over negative x plus the cube root of negative x and let's see, can we simplify this any? Well we could rewrite this
as the negative of one over x and then yeah, I could view negative x as the same thing as negative one times x and so we can factor out, or I should say we could
take the negative one out of the radical. What is the cube root of negative one? Well it's negative one, so we could say minus
one times the cube root or we could just say the
negative of the cube root of x and then we can factor out a negative, so this is going to be equal
to negative of one over x plus the cube root of x, which is equal to the negative of g of x, which is equal to the negative of g of x. And so this is odd, f of negative x is equal
to the negative of f of x, or in this case it's g of x, g of negative x is equal
to the negative of g of x. Let's do the third one. So here we've got h of x and let's just evaluate h of negative x. h of negative x is equal
to two to the negative x plus two to the negative of negative x, which would be two to the positive x. Well this is the same thing
as our original h of x. This is just equal to h of x. You just swap these two terms and so this is clearly even. And then last but not
least, we have j of x, so let's evaluate j of, why did I write y? Let's evaluate j of negative x is equal to negative x
over one minus negative x, which is equal to negative
x over one plus x, and let's see, there's no clear way of factoring out a negative or doing something interesting where I get either back to j of x, or I get to negative j of x, so this one is neither and we're done.