Question

Prove that f is invertible and find (f^-1)'((1)/(2) ?

If `f(x) = cos x` for all x `in` `(0, (pi)/(2))`, prove that f is invertible and find `(f^-1)'((1)/(2))`.

Answer 1

`arccos(1/2)= pi/3`

Explanation:

To know if a relation is invertible we must know the criteria for being invertible. The criteria are as follows:

❈ It must be a function, meaning that each value of `x` maps to only 1 `y` value.

❈ The inverse function must also be a function, meaning that each value of `x` maps to only 1 `y` value.

❈ The inverse function must be a reflection of `y=f(x)` in the line `y=x`. (you can test this using http://desmos.com/calculator)

❈ Must satisfy `f(f^{-1}(x)) = x` and `f^{-1}(f(x)) = x`

First of all, as `f(x) = cosx` is a function that maps to only 1 value for each x value, it is invertible for at least a certain range.

The inverse function of `f(x)=cosx` is `f^{-1}(x)=arccos(x)`, by definition.

However since `cosx` is a repeating function it is a many-to-one function, meaning that several x values will give the same y value. This means that the inverse function will be a one-to-many relation (since y and x switch places), and therefore not a function. This problem can be overcome by restricting the domain, which has already been done: `x in (0, pi/2)`. The function is defined for this domain.

`arccos(1/2)= pi/3`

Now to prove that it is invertible we want to test it using the following equations:

`f(f^{-1}(x)) = x` and `f^{-1}(f(x)) = x`

`cos(arccos(1/2)) =1/2`
`arccos(cos(pi/3)) = pi/3`