Question

`f` is a function on `RR` where `f(x)=|x^2-2|`, How do you find `f^-1(x)`?

Answer 1

Look below

Explanation:

first, we have a detail that the function `f` is present for `RR`.

the function is `\abs{x^2-2}`, so let's make a table from `x >= 0` to `x <= 2`

now graph the points, `(0, 2), (1, 1), (2, 2)`

graph{\abs{x^2-2}[-10, 10, -5, 5]}

now, the inverse of an absolute value function is basically the same function, which is `abs{ x^2-2}`

Answer 2

Explanation:

Let's turn `f(x)=abs(x^2-2)` into a piece-wise function.

Now, we see that `absa=a` when `a>=0`
Similarly, we see that `absa=-a` when `a<0`

Therefore, we have:
`f(x)=x^2-2` when `x^2-2>=0`

`f(x)=-(x^2-2)` when `x^2-2<0`

Let's graph the two parabolas. When we do, we get:

Now, the blue parabola applies when `x^2-2>=0`, and green parabola applies when `x^2-2<0`.

Therefore, we have(focus on the red graph):

Now, to find its inverse, we have to reflect the red graph over the line `y=x`.

Let's try this mathematically.

Our piece-wise function was:
`f(x)=x^2-2` when `x^2-2>=0`

`f(x)=-(x^2-2)` when `x^2-2<0`

Let's find the inverse function for each situation.

Now, remember that `f(f^-1(x))=x` and `f^-1(f(x))=x`

Therefore, we can now find the inverse functions.

`f(x)=x^2-2=>x=(f^-1(x))^2-2`

=>`x+2=(f^-1(x))^2`

=>`+-sqrt (x+2)=f^-1(x)` when `x^2-2>=0`

Similarly,

`f(x)=-x^2+2=>x=-(f^-1(x))^2+2`

=>`x-2=-(f^-1(x))^2`

=>`2-x=(f^-1(x))^2`

=>`+-sqrt (2-x)=f^-1(x)` when `x^2-2<0`

We now graph these two sideways parabolas:

When `y>=sqrt2` or `y<= -sqrt2` , then we apply our first function.

When `-sqrt2<y< sqrt2` , then we apply our second function.

We therefore have:

Just note that `+-sqrt(x+2)` and other relations like this one is not a function but a relation because there are more than one output for a given input.