# composition of funtion

• Sep 29th 2012, 01:22 PM
Candy101
composition of funtion
a little confuse with this question

Find(fogoh)^-1

I dont know How my teacher get it ash^-1(g^-1(f^-1(x)))

I thought it would bef^-1(g^-1(h^-1(x)))

Because if i have Just(fog)^-1(x)i would getf^-1(g(x))

I HAD to find f^-1(x) , g^-1(x) , h^-1(x) along with (fogoh)^-1

F(X)= 3x+4 answer for this is x-4/3
g(x) = 2/x answer is 2/x

I know how to get those i just dont understand (fogoh)^-1 and how
To put it together

He put them together as

h^-1(g^-1(x-4/3))=h^-1(2/(x-4/3)) = h^-1(6/x-4) = (6/x-4)^2 +4

Thank you.
• Sep 29th 2012, 01:34 PM
Plato
Re: composition of funtion
Quote:

Originally Posted by Candy101
a little confuse with this question
Find(fogoh)^-1
I dont know How my teacher get it ash^-1(g^-1(f^-1(x)))

Do you understand that $\displaystyle f^{-1}\circ f=x~?$

Then $\displaystyle (h^{-1}\circ g^{-1}\circ f^{-1})\circ(f\circ g\circ h)=~?$

Work from inside out.
• Sep 29th 2012, 01:39 PM
TheEmptySet
Re: composition of funtion
If you draw a diagram of the domains and ranges it will make the situation clearer on why the order must be reversed.

By the way, in general, this is called the reverse order law.

Rember that the domain of f is the range of f inverse and the range of f is the domian of f inverse.

Attachment 24973
• Sep 30th 2012, 11:44 AM
Candy101
Re: composition of funtion
The "work from inside out" concpet help me to remember it
But im having a lil trouble putting it together.

F^-1= x-4/3
g^-1=2/x
h^-1= x^2+4

h^-1(g^-1(f^-1(x)))

I replace (F^-1) (x) with x-4/3

h^-1(g^-1( x-4/3 ))

Then i use 2/x and for g^-1 and i replace all the x in 2/x witb x-4/3

So this is what i got so far

h^-1(2/(x-4/3))

And the teacher got the same thing
But then got
h^-1 (6/x-4)= (6/x-4)^2+4 i dont understsnd where he got the 6 from. How did he get (6/x-4)= from h^-1(2/(x-4/3))?
• Sep 30th 2012, 12:41 PM
Candy101
Re: composition of funtion
Never mind i got it! Thanks guys for all the help!!!
I can actually do these kind of problems thanks to you!
• Sep 30th 2012, 12:43 PM
Soroban
Re: composition of funtion
Hello, Candy101!

Everyone (including your teacher) is making very hard work . . .

Quote:

$\displaystyle \text{Given: }\:\begin{Bmatrix} f(x) &=& 3x + 4 \\ \\[-4mm] g(x) &=& \dfrac{2}{x} \\ \\[-3mm] h(x) &=& \sqrt{x-4} \end{Bmatrix}$

$\displaystyle \text{Find }\,(f\circ g\circ h)^{-1}$

First find $\displaystyle f\circ g\circ h$, then find its inverse . . .

$\displaystyle f\left(g\big[h(x)\big]\right) \;=\;f\left(g\bog[\sqrt{x-4}\big]\right) \;=\; f\left(\frac{2}{\sqrt{x-4}}\right) \;=\;3\cdot\frac{2}{\sqrt{x-4}} + 4$

We have: .$\displaystyle y \;=\;\frac{6}{\sqrt{x-4}} + 4$

To find the inverse function:
. . Switch x and y; solve for y.

We have: .$\displaystyle x \;=\;\frac{6}{\sqrt{y-4}} + 4$

. . . . . $\displaystyle x - 4 \:=\:\frac{6}{\sqrt{y-4}}$

. . . $\displaystyle \sqrt{y-4} \;=\;\frac{6}{x-4}$

. . . .$\displaystyle y - 4 \;=\;\left(\frac{6}{x-4}\right)^2$

n . . . . . $\displaystyle y \;=\;\left(\frac{6}{x-4}\right)^2 + 4$

Therefore: .$\displaystyle \big(f\circ g\circ h\big)^{-1} \;=\;\left(\frac{6}{x-4}\right)^2 + 4$