Graphs of Inverse Functions
Date: 07/25/2002 at 16:57:07 From: Jonida Subject: Graphs of inverse functions Suppose y=f(x) and y=g(x) are inverses. It turns out that the graph of one function is the graph of the other, but reflected across the line y=x. Explain why this is the case.
Date: 07/26/2002 at 12:43:16 From: Doctor Peterson Subject: Re: Graphs of inverse functions Hi, Jonida. There are several ways to think about this. I'll suggest a couple. First, let's take a graph: y y=f(x) | / | / | / | / | / | / +-------------x The inverse function means that y = f(x) whenever x = g(y) and x = g(y) whenever y = f(x) That means that the point (x,y) is on the graph of y = f(x) whenever the point (x,y) is on the graph of x = g(y). That is, the graph we have drawn is also the graph of x = g(y): y x=g(y) | / | / | / | / | / | / +-------------x But that's not the graph we want; we want to see the graph of y=g(x). How do we get that? We have to swap the variable names: x y=g(x) | / | / | / | / | / | / +-------------y Now we have the right variables, but the axes are in the wrong places. Let's make these two graphs, for y=f(x) and y=g(x). If you've drawn the second graph on sufficiently transparent paper, you can just flip the paper over and place it on top of the first, so that the x and y axes on the two graphs line up together. Essentially what you have done is to rotate the paper around the line y=x: y y=x x y=x | / | / | / | / | / -----> | / | / | / | / | / | / | / +-------------x +-------------y That's the same as reflecting every point on the paper in that line, so our graph of g is now the reflection of the graph of f we started with. Here's a less visual and more analytical way to see it. We'll start the same way: We make the graph: y y=f(x) | / | / | / | / | / | / +-------------x The inverse function means that y = f(x) whenever x = g(y) and y = g(x) whenever x = f(y) That means that the point (a,b) is on the graph of y = f(x) whenever the point (b,a) is on the graph of y = g(x), since b=f(a) means the same thing as a=g(b). So we can look at these two points: y | Q(b,a) / a+---* / | | / | | / b+---/-----*P(a,b) | / | | +---+-----+---x b a Now, what does it mean to say that Q(b,a) is the reflection of P(a,b) in the line y=x? One way you can think of that is that the line segment PQ between the points is (1) perpendicular to y=x, and (2) bisected by y=x. Can you see that? You might want to imagine folding the paper along that line, so that P and Q coincide, and think about what the line PQ will look like. Now to show that PQ is perpendicular to y=x, we need to show that its slope is -1. That's true, right? And to show that the midpoint of PQ is on y=x, we need to find the midpoint and show that its coordinates are equal. You can do that too, right? Then you've proved it! If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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