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BSMSMSTMSPHD Jul17-06 11:03 PM

Finding Inverse Functions
Hi everyone...

I am currently teaching summer Precalculus at the University of New Hampshire, and I have come to the section on inverse functions. I have no problems relating the basic definitions: one-to-one, horizontal line test, etc., but I am looking for clarification on one point.

When we teach students how to "find" inverse functions, the steps are usually the same:

I. Confirm that function y=f(x) is one-to-one.

II. Solve for x (if possible).

III. Switch the variables so that you have y = f-1(x).

So, I have two questions. Please answer them by number.

1. Using precalculus mathematics, can we always perform the second step (solving for x)? I hate to use the words "never" or "impossible" with students unless I really mean it.

As an example, is there a way to write the inverse of y = x^3 + x?

2. Using non-precalculus methods, can we... (same question).


CRGreathouse Jul17-06 11:38 PM

If a function passes the horizontal line test, it has an inverse. That doesn't mean it has a closed-form inverse -- many functions don't. It's beyond precalculus to determine most inverses, but the simple criterion above lets them see if there is one.

The absolute value functon, for example, fails the horizontal line test since |-3|=|3|, for example. It can't have an inverse (unless its new domain is restricted) since 3 would have to map to both 3 and -3, making the 'inverse' not a function (failing the vertical line test).

Office_Shredder Jul18-06 06:44 AM

Even if it fails the horizontal line test, you can usually break off the part of the function that's relevant, and find the inverse for that. For example, there is no inverse for y=x^2. However, there is an inverse for y=x^2 for x > 0 (namely, the square root function).

HallsofIvy Jul18-06 08:33 AM


As an example, is there a way to write the inverse of y = x^3 + x?

Since the derivative of y, 3x2+ 1, is never negative, this is one-to-one and has an inverse.
Finding an inverse is eqivalent to solving an equation. Here the equation is x3+ x- y= 0, to be solved for x for all y. There are 3 solutions to that equation but two of them are complex conjugates and only one real. It is the real solution that would be the inverse. There is a formula (Cardano's formula) for solving cubic equations so you could write out a formula for f-1(x) but it is very complicated and I suspect won't give anything nice here.

In general, even when the function has an inverse, typically that inverse can be written in a simple formula. In fact, often we define a new function to be the inverse. The inverse of f(x)= bx, b>0, (which is one-to-one for all x and has an inverse) is defined to be f-1(x)= log[sub]b[sub](x). The inverse of f(x)= sin(x) (which is not one-to-one but is if we restrict x to between [itex]-\pi[/itex] and [itex]\pi[/itex]) is defined to be f-1(x)= Arcsin(x).

VietDao29 Jul18-06 11:02 AM


Quote by HallsofIvy
...The inverse of f(x)= sin(x) (which is not one-to-one but is if we restrict x to between [itex]-\pi[/itex] and [itex]\pi[/itex]) is defined to be f-1(x)= Arcsin(x).

No, we do not restrict x to between [itex]-\pi[/itex] and [itex]\pi[/itex]. Since, for every value y : -1 < y < 1, we can find more than one [tex]\theta[/tex], such that: [tex]\sin \theta = y[/tex].
We do, however, restrict it to between [tex]- \frac{\pi}{2}[/tex], and [tex]\frac{\pi}{2}[/tex], as the function sin(x) is strictly increasing in that interval.

hypermorphism Jul18-06 01:52 PM

No and Yes
1. It is not always the case that one can find a closed-form expression for the inverse function when an inverse exists. For a simple example, take the function f(x) = x + cos(x). It is obvious that the function is everywhere 1-1 (we pretty much label each cosine with its argument to make each one unique) and thus has an inverse, but it is impossible to solve for that inverse using pre-calculus methods (and the inverse cannot be written in closed form using elementary functions).
2. Using calculus, we can derive the basic theorem that the derivative of [itex]f^{-1}(y)[/itex] is just [itex]\frac{1}{f'(x)}[/itex] where [itex]x = f^{-1}(y)[/itex]. Using this theorem and knowledge of Taylor expansions, we can write a series expansion for [itex]f^{-1}(y)[/itex] centered about any y we choose. We can also resort to a plethora of numerical methods, ie., Newton-Raphson, to approximate specific values.

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