a little confuse with this question
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
h(x)= radical x-4 answer is x^2+4
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
I just dont understnad the (fogoh)^-1 part can u please help me with this
The "work from inside out" concpet help me to remember it
But im having a lil trouble putting it together.
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
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))?
Everyone (including your teacher) is making very hard work . . .
First find , then find its inverse . . .
We have: .
To find the inverse function:
. . Switch x and y; solve for y.
We have: .
. . . . .
. . .
. . . .
n . . . . .