Question

Question #a818a

Answer 1

Please see the explanation.

Explanation:

Here is the graph:

![Desmos.com]()

This looks like a cosine function that has been multiplied by an amplitude and phase shift to the right:

`f(x) = Acos(x - phi) = sin(x) + cos(x)`

My graphing tool allows me to obtain values of points and I can tell you that `A = sqrt(2)`

`cos(x - phi) = 1/sqrt(2)sin(x) + 1/sqrt(2)cos(x)`

This fits the trigonometric identity:

`cos(x - phi) = sin(x)sin(phi) + cos(x)cos(phi)`

where `cos(phi) = sin(phi) = 1/sqrt(2)`

This happens at `phi = pi/4`

The graphing tool confirms that the x coordinates are shift by `pi/4`.

`f(x) = sqrt(2)cos(x - pi/4)`

Substitute `f^-1(x)` for every x:

`f(f^-1(x)) = sqrt(2)cos(f^-1(x) - pi/4)`

The left side becomes x by definition:

`x = sqrt(2)cos(f^-1(x) - pi/4)`

Make the `sqrt(2)` on the right disappear by multiplying both sides by `sqrt(2)/2`:

`sqrt(2)/2x = cos(f^-1(x) - pi/4)`

Use the inverse cosine on both sides:

`cos^-1(sqrt(2)/2x) = f^-1(x) - pi/4`

Solve for `f^-1(x)`:

`f^-1(x) = cos^-1(sqrt(2)/2x) + pi/4`

To confirm that this is truly an inverse, verify that `f(f^-1(x)) = f^-1(f(x)) = x`