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Symmetry of Graphs: Odd and Even Functions - Problem 2
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We’re proving a function is even or odd and I want to start with this function; f(x) equals x minus 4 over x plus 4. Remember that, the strategy here is to plug in –x and see what happens to the expression. If I plug –x in, I get –x minus 4, over –x plus 4. Now I can actually cancel a negative over the numerator and denominator, by multiplying the top and bottom by -1. Let me do that.

That will give me +x plus 4 on top and on the bottom it will give me +x minus 4. This is exactly the reciprocal of this, which is neither the opposite of this nor is it equal to this. So this function is neither even nor odd because the result was neither f(x) nor –f(x).

Let’s try another example. G(x) equals 2 to the x minus -1 over 2 to the x plus 1. It looks a lot like what we just did, but let’s treat it as a brand new function. Let’s see what happens. I plug in –x and I get 2 to the –x minus 1 over, 2 to the –x plus 1. Now I can treat these two to the –x as if they were 1 over 2 to the x and get rid of them by multiplying the top and bottom by 2 to the x.

Let me distribute this over the top and bottom. I get 2 to the x times 2 to the –x which is 1, minus 2 to the x over 1, plus 2 to the x. I can also factor a -1 out of the top and get minus and I can actually reverse the order of this. 2 to the x minus 1 over 2 to the x plus 1. This is exactly the opposite of my original function.

That means that g(x) is an odd function. G(-x) equals g(x) and that’s the very definition of an odd function. G is odd.

Again when you’re proving whether a function is even or odd, start by plugging in –x and using algebra. See if you can manipulate it into either minus g(x) or g(x).

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