Inverse functions

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pope4

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One of my homework questions asks to write an inverse function of y=5x+2, and it got me thinking about the exact meaning behind the term. I'm confused because the answer came out to be y=1/5(x-2), but no matter what definition I employ it just doesn't add up for me. On one hand, the inverse of 2 would be -2 because 2+(-2)=0, but on the other 5^-1=1/5 and 1/5*5=1. Why does it work that way? Why is one way of taking the inverse appropriate for half the equation and another for the other half?
 
pope4 said:
One of my homework questions asks to write an inverse function of y=5x+2, and it got me thinking about the exact meaning behind the term. I'm confused because the answer came out to be y=1/5(x-2), but no matter what definition I employ it just doesn't add up for me. On one hand, the inverse of 2 would be -2 because 2+(-2)=0, but on the other 5^-1=1/5 and 1/5*5=1. Why does it work that way? Why is one way of taking the inverse appropriate for half the equation and another for the other half?
Click to expand...
The word "inverse" is used to describe several different ways in which something can be "undone".

The additive inverse of 2 is -2, which undoes addition of 2. The multiplicative inverse of 5 is 1/5, which undoes multiplication by 5. The question is about the inverse function, which we write as [imath]f^{-1}[/imath] by analogy with the multiplicative inverse, but which is not the same thing.

Did you carefully read the definition of an inverse function? A function [imath]g[/imath] is the inverse of a function [imath]f[/imath] if [imath]f(g(x))=x[/imath] for all x, and [imath]g(f(x))=x[/imath]. That is, each function undoes the other when you combine them with composition. (So we could call [imath]g[/imath] the "compositive inverse" of [imath]f[/imath], I suppose, though I've never said that in my life!)

Now, it happens that the inverse of your function undoes both a multiplication and an addition, so it involves an additive inverse and a multiplicative inverse. That, I think, is what you are seeing.

By the way, this is just what you are doing when you solve an equation like [imath]5x + 2 = 17[/imath]. You first undo the addition, by subtracting, and then undo the multiplication, by dividing. That's exactly what the inverse function does. So it really does all fit together.
 
Dr.Peterson said:
The word "inverse" is used to describe several different ways in which something can be "undone".

The additive inverse of 2 is -2, which undoes addition of 2. The multiplicative inverse of 5 is 1/5, which undoes multiplication by 5. The question is about the inverse function, which we write as [imath]f^{-1}[/imath] by analogy with the multiplicative inverse, but which is not the same thing.

Did you carefully read the definition of an inverse function? A function [imath]g[/imath] is the inverse of a function [imath]f[/imath] if [imath]f(g(x))=x[/imath] for all x, and [imath]g(f(x))=x[/imath]. That is, each function undoes the other when you combine them with composition. (So we could call [imath]g[/imath] the "compositive inverse" of [imath]f[/imath], I suppose, though I've never said that in my life!)

Now, it happens that the inverse of your function undoes both a multiplication and an addition, so it involves an additive inverse and a multiplicative inverse. That, I think, is what you are seeing.

By the way, this is just what you are doing when you solve an equation like [imath]5x + 2 = 17[/imath]. You first undo the addition, by subtracting, and then undo the multiplication, by dividing. That's exactly what the inverse function does. So it really does all fit together.
Click to expand...

It makes much more sense now, thank you!!
 
@pope4

I like to think of a function as a machine where you have an input number and an output number.

In the case of the function \(\displaystyle y= 5x+2\), it takes a number, multiplies it by 5 and then adds 2.

For example, if you input the number 3, it will output the number 17.

The inverse function can be thought of as another machine which "undoes" what the function has done.

In this case, the machine needs to subtract 2 and then divide by 5 ie \(\displaystyle y = \frac{1}{5}(x-2)\).

So, if you input the number 17, it will output the number 3.
 
It once bothered me why the process of how you were to deduce the inverse function of f(x) worked.

Suppose we have [imath]f(x) = \text {some expression in } x[/imath].

Now let [imath]y = f(x) \implies y = \text {some expression in } x[/imath].

We solve for x in terms of y. That gives us [imath]x = \text {some expression in terms of } y.[/imath]

If, and only if, that expression in terms of y defines a function,
we can say [imath]g(y) = \text {expression in terms of } y \implies x = g(y).[/imath]

But we have[math]y = f(x) \text { and } g(y) = x \implies \\ g(f(x)) = x \implies g(x) \text { is the inverse function of } f(x) \text { by definition.}[/math]
 
You have a number x, then you multiply it by 5 and then you add 2.
How would you retrieve the number x? Well if you subtract 2, you'll have 5x. Now you divide by 5 and you have x. What exactly did I do again? Ah, I subtracted 2 and then divided by 5. So define g(x) to be that function, ie let g(x) = (x-2)/5.
Now if f(x) = 5x+2, then the inverse of f(x) will be g(x) = (x-2)/5.
 
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