Question

Help?explain

Answer 1

Restricting the domain of `f(x)` to `[0, oo)`

`f^(-1)(y) = ln(y+sqrt(y^2-1))`

which has domain `[1, oo)`

Explanation:

Given:

`f(x) = 1/2(e^x+e^(-x))`

(Side note: This is the definition of the function `cosh x`)

Note that `f(x)` is not one to one. For example:

`f(-1) = 1/2(e^(-1)+e^1) = 1/2(e^1+e^(-1)) = f(1)`

In fact `f(-x) = f(x)` for all `x in RR`

Since `f(x)` is not one to one, its inverse function cannot be well defined without restricting the domain of `f(x)`.

If we restrict the domain to non-negative numbers, then it does have an inverse. That is we restrict the domain to `x >= k` where `k = 0`.

We cannot make `k` any less, since `f(x)` takes the same values for negative arguments as for their opposites.

Let:

`y = f(x) = 1/2(e^x+e^(-x))`

Then:

`e^x-2y+e^(-x) = 0`

Multiplying through by `e^x`, we have:

`0 = (e^x)^2-2y(e^x)+1`

`color(white)(0) = (e^x)^2-2y(e^x)+y^2-(y^2-1)`

`color(white)(0) = (e^x-y)^2-(sqrt(y^2-1))^2`

`color(white)(0) = ((e^x-y)-sqrt(y^2-1))((e^x-y)+sqrt(y^2-1))`

`color(white)(0) = (e^x-y-sqrt(y^2-1))(e^x-y+sqrt(y^2-1))`

Hence:

`e^x = y+-sqrt(y^2-1)`

Taking natural logarithms of both sides we find:

`x = ln(y+-sqrt(y^2-1))`

The positive value of `x` corresponds to taking the `+` sign in the radicand (note that the negative sign would give `y-sqrt(y^2-1) < 1` which has negative logarithm).

So:

`x = ln(y+sqrt(y^2-1))`

and our inverse function is:

`f^(-1)(y) = ln(y+sqrt(y^2-1))`

(This is `cosh^(-1)(y)`)

This has domain `[1, oo)`, which is the range of `f(x)`.