Question

How do you find the inverse of `f(x) = x^2 + 2` and is it a function?

Answer 1

`bar(f)(x)=+-sqrt(x-2)`
`color(white)("XXX")`It is not a function (since it generates more than one solution for single values of `x`.

Explanation:

If `color(red)(bar(f)(x))` is the inverse of `color(blue)(f(x))`
then by definition of inverse `color(blue)(f(color(red)(bar(f)(x))))=color(green)(x)`

and if `color(blue)(f(color(black)(x)))=color(black)(x)^2+2`
then
`color(white)("XXX")color(blue)(f(color(red)(bar(f)(x))))=color(red)(bar(f)(x))^2+2`

Therefore
`color(white)("XXX")color(red)(bar(f)(x))^2+2=color(green)(x)`

`color(white)("XXX")color(red)(bar(f)(x))^2=color(green)(x)-2`

`color(white)("XXX")color(red)(bar(f)(x))=+-sqrt(color(green)(x)-2)`

Answer 2

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

Explanation:

The equation `f=x^2+2` represents the parabola, with vertex V(0, 2) and axis along y-axis, in the positive direction, from V. Note that, if (x, f) is on the parabola), so is `(-x, f)`. `xtof` is 2-1 mapping.

So, the inverse obtained by solving for x is
`f^(-1)f(x)=x=+-sqrt(f-2)`

The first equation represents the half of the parabola in the 1st quadrant and the second is for the other half, in the second quadrant.

f is single valued but the inverse is double-valued.
Inverse trigonometric functions are many-valued, returning a set of values. .
.