# When Does An Inverse Function Exist?

• Aug 25th 2009, 11:36 AM
saberteeth
When Does An Inverse Function Exist?
I have understood that an inverse of a function exists when the relation is one to one. It is clear to understand when the function is a whole number. But how do you apply that rule to an equation? What are the steps involved?

f(x) = 3x - 4 the inverse of the function exists
f(x) = (2x - 3) / (x -7); x is not equal to 7 the inverse of the function does not exist
f(x) = (x^2) - 1 the inverse of the function does not exist

What are the steps in between?
• Aug 25th 2009, 12:49 PM
Matt Westwood
If you write it in the form $y = 3x + 4$ you may find it's easier to see what's going on.

What you do is see whether you can get it in the form where $x$ is on its own on one side and there's a complicated manipulation of $y$ that you can do to get to $x]$.

So you have:
$y = 3x + 4$
Subtract 4 from both sides:
$y - 4 = 3x$
Divide both sides by 3
$\frac {y-4} 3 = x$
which can be written $\frac y 3 - \frac 4 3$

and it is clear that $x$ is a function of $y$.

The next one's trickier:
$y = \frac{2x - 3} {x - 7}$
Multiply by $x - 7$:
$y (x-7) = 2x - 3$
$xy-7y = 2x - 3$
$xy - 2x = 7y - 3$
$x (y - 2) = 7y - 3$
$x = \frac {7y - 3} {y-2}$

The only reason that's not a function is because it's not defined at $y=2$, and so is not one-to-one. There's no image of 2.

Can you do the third one now?
• Aug 25th 2009, 11:12 PM
saberteeth
Quote:

Originally Posted by Matt Westwood
If you write it in the form $y = 3x + 4$ you may find it's easier to see what's going on.

What you do is see whether you can get it in the form where $x$ is on its own on one side and there's a complicated manipulation of $y$ that you can do to get to $x]$.

So you have:
$y = 3x + 4$
Subtract 4 from both sides:
$y - 4 = 3x$
Divide both sides by 3
$\frac {y-4} 3 = x$
which can be written $\frac y 3 - \frac 4 3$

and it is clear that $x$ is a function of $y$.

The next one's trickier:
$y = \frac{2x - 3} {x - 7}$
Multiply by $x - 7$:
$y (x-7) = 2x - 3$
$xy-7y = 2x - 3$
$xy - 2x = 7y - 3$
$x (y - 2) = 7y - 3$
$x = \frac {7y - 3} {y-2}$

The only reason that's not a function is because it's not defined at $y=2$, and so is not one-to-one. There's no image of 2.

Can you do the third one now?

Ok. For (x^2) - 1 . The output will be x = square root of (y + 1) which will give a positive as well as a negative answer. So, it's inverse function will not exist.. Also, for http://www.mathhelpforum.com/math-he...101fea40-1.gif, after simplifying it, if we put y = 2 in the denominator of http://www.mathhelpforum.com/math-he...2b2b4768-1.gif, the answer will be infinite ( i. e not defined) hence, this is the only reason the inverse function will not exist.. Am i right? (Thinking)
• Aug 26th 2009, 12:10 PM
Matt Westwood
You got it. Nice one.
• Aug 26th 2009, 12:21 PM
saberteeth
Thanks much(Happy)
• Aug 26th 2009, 05:01 PM
math84
I don't understand why the second one has to hit 2 unless we are given that the codomain includes all real numbers. What am I missing here?
• Aug 26th 2009, 07:08 PM
saberteeth
Quote:

Originally Posted by math84
I don't understand why the second one has to hit 2 unless we are given that the codomain includes all real numbers. What am I missing here?

Hi there! The reason is:

In elementary mathematics, the domain is often assumed to be the real numbers, if not otherwise specified, and the range is assumed to be the image.

Source: Inverse function - Wikipedia, the free encyclopedia
• Aug 26th 2009, 07:33 PM
QM deFuturo
Quote:

Originally Posted by saberteeth
Hi there! The reason is:

In elementary mathematics, the domain is often assumed to be the real numbers, if not otherwise specified, and the range is assumed to be the image.

Source: Inverse function - Wikipedia, the free encyclopedia

I know what a Domain and Range are. What is an "image" in this context??
• Aug 26th 2009, 07:36 PM
math84
Quote:

Originally Posted by saberteeth
Hi there! The reason is:

In elementary mathematics, the domain is often assumed to be the real numbers, if not otherwise specified, and the range is assumed to be the image.

Source: Inverse function - Wikipedia, the free encyclopedia

Do you mean the codomain is assumed to be the range? I don't understand.(Headbang)

There is no need for 2 to have an image under the inverse unless it's in the domain of the inverse, which is only going to need to happen it is in the codomain of our original function. Unless I'm mistaken, 2 is NOT in the range of our original function, so it's probably not in the assumed codomain of our original function either. However, I can't see any other way for the inverse to not exist since the function is apparently injective, so I believe you are correct. I'm missing something here.