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.

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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

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))?

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!

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