Question about inverse function

Question

As a high school student, I have learnt about inverse functions, but not its theorem (inverse function theorem). I looked at wikipedia but the mathematics is too hard for me to understand. Can somebody explain in simple terms the conditions required for a function to have an inverse function? That is, if $y$ is a function of $x$, under what conditions should $x$ be a function of $y$ ?

Edit:

If we know that $y$ is not a function of $x$, then is it necessary that $x$ is also not a function of $y$? How can we prove that?

Answer 1

Some preliminaries:

Note that a function $f(x)$ is one-one if no two values of $x$ produce the same value of $f(x)$, that is, $$f(x_1) \neq f(x_2) \, \,,\,\, x_1 \neq x_2$$

Note that a function $f: A \to B$ is onto if the range of $f$ is $B$. In other words, if for each $b \in B$, there exists atleast one $a \in A$ such that $$f(a)=b$$ it is onto.

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Now, note that bijectivity of a function (being both one-one and onto) is a necessary and sufficient condition to have an inverse over the whole domain and range. For more, you can see here.

Answer 2

Graphically, if $f$ is a continuous function it will have an inverse if the function $f$ passes the so-called horizontal line test. Here a horizontal line must intersect the curve given by the function $f$ once only.

If your function fails the horizontal line test it may still be possible to define an inverse for your function $f$ by restricting the domain of the function $f$ in such a way so as to now pass the horizontal line test.

As an example of this consider the function $f(x) = x^2$ which is a parabola centred at the origin. When $y > 0$ it clearly fails the horizontal line test as a horizontal line intersects the curve twice. However, if one restricts the domain of the original function $f$ to $x \geqslant 0$, a horizontal line will now only intersect the curve once. The function $f$ thus has an inverse. In this case it can actually be found and is given by $f^{-1} (x) = \sqrt{x}$.

Alternatively, one may just as easily restrict the domain of the function $f$ to $x \leqslant 0$. In this case the function with this restrict domain again satisfies the horizontal line test meaning it has an inverse. In this case $f^{-1} (x) = -\sqrt{x}$.

Answer 3

A function $f $ is a mapping or a pairing from one set $A $ called a domain to another set $B $, so that for each distinct element $x\in A $ there is a specific $y_x = f (x)\in B $. That is if $x $ is $x $ then $f (x) $ will always be the same thing. It can't be that $5$ gets mapped to $7$ one time and then $5$ gets mapped to $37$ the next.

There must be a $y_x \in B$ for every didstinct $x\in A $ but the $y_x $ need not be distinct. It's perfectly possible (quite probable ) that $f (w)=f (x) $ but $x\ne w $.

But some functions are "one to one" so that indeed for every $y\in B $ there is only one $x\in A $ where $f (x)=y $.

Example $f (x)=x^2 -2x+1$ is not one to one because $f (3)=f (-1)=4$. But $g (x)=7x+2$ is because whenever $g (x)=g (w)$ then $7x+2=7w+2$ which means $x=w $.

If a function is one to one then that means the mapping "goes both ways". The function $y=f (x) $ is a function that maps $A $ to $B $ but because all the $f (x) $ are distinct and unique the $x=f^{inv}(y) $ is a function that maps $B $ to $A $.

This isn't a theorem. It's just a definition.

Now I said not all functions are one to one so they don't all have inverses. But if a function is continuous, and if at some point the function is decreasing or increasing at the point, then for some interval around to point there is an area (possibly very small) where a portion of the function is one to one and therefore has an inverse in that region.

This is common sense if you draw it. If the function is decreasing, for example, then it will continue to decrease for a while and in that interval while it is decreasing all the points it maps to will be distinct because it's not increasing to map any of them twice. So for that moment in which it is decreasing the function has an inverse.

Well, that's an intuitive folksy argument-- it isn't a proof. And that is half of what the inverse function theorem says.

The other half?

Okay. Calculus is all about measuring the slopes of functions at certain points. Now, functions (that are not lines) will have slopes that change all the time. Calculus cares how these slopes behave (because the slopes tell as how fast the function is changing at a given point which allows us to predict the behavior of the function...VERY useful.) We call the slope of a function "the derivative".

The second half of the inverse function theory is that if a function has a derivative of $s_x $ at the point $y=f (x) $. Then the inverse function (in the small interval where the function has an inverse) will have a derivative of $\frac 1 {s_x} $ at the point $x =f^{inv}(y) $.

This also makes sense if you draw it. If $f $ maps points of $A $ to points of $B $, then $f^{inv} $ maps the same points of $B $ back to $A $. So their graphs are going to be exactly the same but with the $x $ and $y $ axes flipped! So that means their slopes are going to be the same proportions but with the axes flipped. So they'll all be multiplicative recipricals.

And that is the inverse function in a nutshell.