Inverse Functions

[Graph of line y=2x-5 and its inverse y=(x+5)/2]

[Graph of exponential e^x and its inverse ln(x)]

Some Functions and their Inverses

An invertible function is a function that can be inverted. An invertible function must satisfy the condition that each element in the domain corresponds to one distinct element that no other element in the domain corresponds to. That is, all of the elements in the domain and range are paired-up in monogomous relationships - each element in the domain pairs to only one element in the range and each element in the range pairs to only one element in the domain. Thus, the inverse of a function is a function that looks at this relationship from the other viewpoint. So, for all elements a in the domain of f(x), the inverse of f(x) (notation: f^-1(x)) satisfies:
f(a)=b implies f^-1(b)=a
And, if you do the slightest bit of manipulation, you find that:
f^-1(f(a))=a
Yielding the identity function for all inputs in the domain.

When we graph functions and their inverses, we find that they mirror along the line x=y. This is only logical. From our definition, we know that for each (a,b) in f(x) there will be a (b,a) in f^-1(x):

As an exercise, draw a graph of any one-to-one function you know - or make one up. Then, take some points (a,f(a)) and plot them at (f(a),a) in a different color. This second curve is the inverse of the first.

Lines can be easily manipulated to find inverses. Except for horizontal and vertical lines, each point x on a line corresponds to one y. I will illustrate the general procedure for finding the inverse of a function with a line. Let us take the line described by y = 2x-5.
General procedure for finding the inverse of a function:

Interchange the variables - First, we will exchange the variables. We do this because we want to find the function that goes the other way, by mapping the old range onto the old domain. So our new equation is x=2y-5.
Solve for y -The rest is simply solving for the new y, which gives us:
2y-5 = x 2y = x+5 y = (x+5)/2 Hence, y^-1(x) = (x+5)/2

Now that that is out of the way, we have bigger inversion problems to worry about. What happens if we try to find the inverse of a parabola? Well, look at the graph:

What happens is that, because a parabola is not a one-to-one the inverse can't exist because for various values of x (all x>0) f^-1(x) has to take on two values! To solve this problem in taking inverses, in many cases, people decide to simply limit the domain. For instance, by limiting the domain of the parabola y=x² to values of x>0, we can say that the function's inverse is y=+sqrt(x). (sqrt(x) means the square root of x or x^1/2) This is done to let the trigonometric functions have inverses.
[Overlayed Graphs of sin, Sin, and ArcSin]

[Overlayed Graphs of sin, Sin, and ArcSin]

As you can see, we can't take the inverse of sin(x) because it is not a one-to-one function. However, we can take the inverse of a subset of sin(x) with the domain of -/2 to /2. The new function inverse we get is called Sin^-1(x) or ArcSin(x).

Inverse Fuction Domain Range

Sin^-1(x) {x: -1 <= x <= 1} -/2 <= f(x) <= /2

Cos^-1(x) {x: -1 <= x <= 1} 0 <= f(x) <=

Tan^-1(x) {x: -infinity <= x <= infinity} -/2 <= f(x) <= /2

Cot^-1(x) {x: -infinity <= x <= infinity} 0 <= f(x) <=

Sec^-1(x) {x: |x| >= 1} 0 <= f(x) <= , f(x) not = /2

Csc^-1(x) {x: |x| >= 1} 0 < |f(x)| <= /2

-David

Inverse Fuction	Domain	Range
`Sin^-1(x)`	`{x: -1 <= x <= 1}`	`-/2 <= f(x) <= /2`
`Cos^-1(x)`	`{x: -1 <= x <= 1}`	`0 <= f(x) <=`
`Tan^-1(x)`	`{x: -infinity <= x <= infinity}`	`-/2 <= f(x) <= /2`
`Cot^-1(x)`	`{x: -infinity <= x <= infinity}`	`0 <= f(x) <=`
`Sec^-1(x)`	`{x: \|x\| >= 1}`	`0 <= f(x) <= , f(x) not = /2`
`Csc^-1(x)`	`{x: \|x\| >= 1}`	`0 < \|f(x)\| <= /2`