• Sep 13th 2011, 01:18 PM
Ok i know how to solve compositions when Im provided with x

Here I am dealing with f of g of h

f(x) = 2x-2
g(x) = x^2
h(x) = 9-x

so its f(g(h(x)

• Sep 13th 2011, 01:23 PM
Plato
Quote:

Here I am dealing with f of g of h
f(x) = 2x-2
g(x) = x^2
h(x) = 9-x
so its f(g(h(x))

$f \circ g \circ h(x) = f\left( {g\left( {h(x)} \right)} \right) = 2(9 - x)^2 - 2$
• Sep 13th 2011, 01:34 PM
Oh I see haha, guess that was a stupid mistake on my part. Guess I just had to leave X as X...
• Sep 13th 2011, 01:39 PM
f(x) = √x
g(x) = 7√1-x

I tried √7-x, √x, √7x they were all wrong
• Sep 13th 2011, 01:55 PM
HallsofIvy
Quote:

f(x) = √x
g(x) = 7√1-x

I tried √7-x, √x, √7x they were all wrong

I don't know what you meant by "tried". Plato told you how to find the compositions.
$f\circ g(x)= f(g(x))= f(7\sqrt{1-x}}= \sqt{7\sqrt{1-x}}$
which could also be written
$\sqrt{7}\sqrt[4]{1- x}$

$g\circ f(x)= g(f(x))= g(\sqrt{x})= 7\sqrt{1- \sqrt{x}}$
• Sep 13th 2011, 02:10 PM
When I submitted both of those, it was marked as wrong...
even after simplfying it down to

√7-x and 7-√x
• Sep 13th 2011, 02:15 PM
Plato
Quote:

When I submitted both of those, it was marked as wrong...
even after simplfying it down to

√7-x and 7-√x

$\sqrt{7\sqrt{1-x}}$
• Sep 13th 2011, 02:50 PM
Nope, still didn't work... I am out of submissions now :(
• Sep 13th 2011, 04:17 PM
Quote:

Nope, still didn't work... I am out of submissions now :(

I am now stuck on G o G of the same problem

I get 7√1-7√1-x which should just be 7√1-x no?
• Sep 13th 2011, 04:28 PM
Plato
Quote:

I am now stuck on G o G of the same problem

I get 7√1-7√1-x which should just be 7√1-x no?

I think that there must be a flaw in what program you are using.

If this is indeed correct $g(x)=7\sqrt{1-x}$

then $g\circ g(x)=7\sqrt{1-7\sqrt{1-x}~}$.

If the program does not take that, then either you are typing the symbols incorrectly or there is a flaw in the program.
• Sep 13th 2011, 05:39 PM
Quote:

Originally Posted by Plato
I think that there must be a flaw in what program you are using.

If this is indeed correct $g(x)=7\sqrt{1-x}$

then $g\circ g(x)=7\sqrt{1-7\sqrt{1-x}~}$.

If the program does not take that, then either you are typing the symbols incorrectly or there is a flaw in the program.

Nope it is showing up as incorrect still
• Sep 13th 2011, 05:56 PM
Plato
Quote:

Nope it is showing up as incorrect still

Then that is a totally incompetent program or you do not know how to use it. I suspect that you are not using enough grouping symbols.
Look at my LaTeX code:
7\sqrt{1-7\sqrt{1-x}} this gives $7\sqrt{1-7\sqrt{1-x}}$.
Note how carefully the various parts are grouped together using grouping symbols.

Does that program allow LaTeX coding?
If so the you can copy and paste from here.

In any case, you need to contact the program administrator to discuss these concerns.
• Sep 13th 2011, 06:30 PM