Question

How to solve Inverse functions

Answer 1

Inverse functions are always reflected about the line y = x; that is, the x and y-ordinates are swapped.

Let's take a point first, (0, 1).
For the inverse point, we need to reflect this point about the line y = x.
In other words, the two points will be swapped.

Thus, the inverse will be (1, 0)

Let's take it one step further, y = 3x + 1
Now, we know there will be an inverse, because the original is a function itself.

In order to solve inverse functions, we have two steps to consider.

Step 1) Interchange the x and y-values around.
Using our original example:


`y = 3x + 1` becomes:
`x = 3y + 1`

Step 2) Solve for y.
All we need to do is make y the subject.

`x - 1 = 3y`

`y = (x - 1)/(3)`

Seems easy right now, but what about equations that are not functions? An example is a circle. A full circle won't have an inverse, but it will if we consider a portion of it.

If we take √(1 - x²), which is the equation of a semicircle, we can't get an inverse because the original is not a function. In this instance, we need to cut the semi-circle such that it is a function. This is called "Restricting the Domain", something that we'll touch when we get to inverse trigonometric functions.

Now, what happens if we take the positive side of the semicircle? We'd simply get a portion of the circle.

So, we'll still have
`y = \sqrt{1 - x^(2)}` as our function, but the inverse only applies for when
`0 \leq x \leq 1` and only in that domain.

We do the same approach as before, interchange and solve.


`x = \sqrt{1 - y^(2)}`

`x^(2) = 1 - y^(2)`

`y^(2) = 1 - x^(2)`

`y = \sqrt{1 - x^(2)}`,
`0 \leq x \leq 1`

Thus, the inverse is itself!

Now, we'll go onto the harder kind of inverse functions, trigonometric functions.
This is hard because of the oscillation that sine and cosine waves have.

The sine wave has a period of 2π, but we only want the functional segment of the wave. Let's take the sine wave from
`-(pi)/(2) \leq x \leq (pi)/(2).`

This allows the sine wave to be a function. Now, if we reflect this about the y = x line, the domain and range will change. In other words, the domain of the sine wave will become the range, and the range of the sine wave will become the domain. Refer to the first picture attached.

Now, similar to the sine wave, the cosine wave also has a period of 2π, but we need to restrict the domain so we can have a function. Remember that we need to keep as much of the original as possible, so our best bet is to take it from:

`0 \leq x \leq \pi`

Our range is the same, so our domain of cosine becomes the range of our inverse, and our range of cosine becomes the domain of our inverse.

Lastly, we have the tangent wave. This is easier because of the asymptotes, so we only need to take from:

`-(\pi)/(2) < x < (\pi)/(2)`.
Refer to the second picture attached.

Because the range of our tangent graph is all real y-values, then the inverse tangent domain becomes all real x.
Refer to the last picture attached.