# Math Help - Inverse functions f-1(f(x)) where x is not in the domain

1. ## Inverse functions f-1(f(x)) where x is not in the domain

Just a quick question because I can't quite seem to figure this out.
They give you the equation f(x) = 1/6 [x^2 -4x +24].
They then ask you to give the inverse function f-1(x) for which the domain is positive.
It is then given that N is a negative real number not in the domain of f-1(x), asking you to find f-1(f(x)). I sub in the value for f(N) into f(x) then sub that back into f-1(f(x)) but it obviously just gives the number N :S.
Just wondering how people here would approach this question.
The answers give the answer as 4-N :S.
Thanks in advance .

2. ## Re: Inverse functions f-1(f(x)) where x is not in the domain

Well first thing is to find where the domain of f is positve, while here, i'd find the range as well.

Then find the inverse of f.

Have a go at that.

3. ## Re: Inverse functions f-1(f(x)) where x is not in the domain

Originally Posted by pickslides
Well first thing is to find where the domain of f is positve, while here, i'd find the range as well.

Then find the inverse of f.

Have a go at that.
Well I managed to get the domain and range with f-1(x), x>=10/3, y>=2.
the inverse function comes out f-1(x) = 2 + (6x - 20)^1/2.
Both intersect at 4,4 and 6,6.
the problem is that subbing in f(n) into f-1(n) just gives n, but the question asks for n being a negative number not in the domain, and it's answer isn't just n :S.

4. ## Re: Inverse functions f-1(f(x)) where x is not in the domain

Originally Posted by tomdoml
Well I managed to get the domain and range with f-1(x), x>=10/3, y>=2.
the inverse function comes out f-1(x) = 2 + (6x - 20)^1/2.
Both intersect at 4,4 and 6,6.
the problem is that subbing in f(n) into f-1(n) just gives n, but the question asks for n being a negative number not in the domain, and it's answer isn't just n :S.
The inverse function of $y= \frac{x^{2}-4\ x +24}{6}$ is...

$x= 2 \pm \sqrt{6\ y -20}$ (1)

... so that it is a multivalued function. If x and y are vinculated to be real variables, then (1) is defined for $y>\frac{10}{3}$. If in general x and y are complex variables, then (1) is defined for all values of y...

Kind regards

$\chi$ $\sigma$

5. ## Re: Inverse functions f-1(f(x)) where x is not in the domain

Since $f(x)$ and $f^{-1}(x)$ are mutually inverse, their graphs are symmetric w.r.t. the diagonal line y = x. Suppose that A has the coordinates (N,0). Calculate the coordinates of B, C, D, E and F taking into account that the red parabola is symmetric w.r.t. x = 2 and CD = DE. The y-coordinate of F is $f^{-1}(f(N))$.