Longtime lurker here, composite function problem

1. Mar 9, 2015

Poset

1. The problem statement, all variables and given/known data

Let f(g(h(x))) = 1/(2-x)

Find g(x) if:

f(x) = (x^2) - 1
h(x) = 3x+1

2. The attempt at a solution

This is what I have:

g(h(x))^2 -1 = 1/(2-x)
g(h(x) = sqrt((3-x)/(2-x))

I'm not sure how to get the h(x) out of this to leave me with just g(x). Please point me in the right direction.

Thank you!

2. Mar 9, 2015

pasmith

The obvious thing to do is to set $y = h(x)$ and find $g(y) = g(h(x))$ in terms of $y$.

3. Mar 9, 2015

Poset

Thank you for your reply. Am I right to assume that the calculations to g(h(x)) I did are correct? And what you are implying is that to get g(x) I need to go backwards from h(x) using the inverse of g(h(x)) and y = h(x) = 3x+1? I'm sorry it's not that obvious to me.

4. Mar 9, 2015

fourier jr

Your second line isn't far off. Then if you replace x there with y=h-1 you'll get g(y).

5. Mar 9, 2015

PeroK

You've got:

$g(h(x)) = \sqrt{\frac{3-x}{2-x}}$

Why not take one more step?

$g(3x+1) = \sqrt{\frac{3-x}{2-x}}$

Can you see what to do to finish things off?

6. Mar 9, 2015

fourier jr

On your last step I see you substituted x with y=3x+1 on the left-hand side, but not the right-hand side. In other words it should be 3-y (etc). Then you should have g(y) but you can set y=x to get g(x). To check your answer calculate f(g(h(x))) & see what you get.

7. Mar 9, 2015

Poset

Nice I just got this. Thank you all for your help

g(x) = sqrt((10-x)/(7-x))

I spent this entire time trying to figure this out. I really couldn't find any similar examples on the internet.

8. Mar 9, 2015

Ray Vickson

Do it systematically: $f(g(h(x)) = g(h(x))^2 - 1$ and $g(h(x)) = g(3x+1)$, so $f(g(h(x)) = g(3x-1)^2 - 1$.

9. Mar 9, 2015

SammyS

Staff Emeritus

Two things to point out to Poset:

First of all; f(x) is not a 1 to 1 function.

So, when you find $g(h(x)) = \sqrt{\frac{3-x}{2-x}}$ , you should include a ± , so you can choose either sign in finding g(x). Thus the choice for g(x) is not unique.

$\displaystyle g(h(x)) = \pm\sqrt{\frac{3-x}{2-x}}$​

Second: The composite function, $\ f\circ g\circ h\$, may have a more restricted domain than the implicit domain of $\ \displaystyle k(x)=\frac{1}{2-x}\$.

Added in Edit: I've been working on this while the recent posts have come in.

Note: $\displaystyle\ \frac{10-x}{7-x}=\frac{x-10}{x-7}$

Last edited: Mar 9, 2015
10. Mar 10, 2015

SammyS

Staff Emeritus
I thought this might be of interest.

Below is a graph of the composite function y = f(g(h(x))), (in black). The range is restricted to y ≥ -1, because the final function, f, has a range, [-1,∞) .

The missing piece of the graph for y =1/(2-x) is shown dotted in grey.

.

11. Mar 10, 2015

Ray Vickson

Just as a matter of interest: what package did you use to make the graph?

12. Mar 10, 2015

SammyS

Staff Emeritus
Ray,

It's called Graph - appropriately enough.

It's not perfect, but usually gets the job done.

13. Mar 12, 2015

Ray Vickson

Thank you for that.