Given f(x)=2x and g(x)=4x-3show that f^-1(g^-1(x))=g(f(x))^-1?

Question

Given f(x)=2x and g(x)=4x-3

show that f^-1(g^-1(x))=g(f(x))^-1

this involves composite functions

Answer 1

F(x) = 2x
Let f(x) =x and x = f^(x)-1
Therefore x = 2f^(x)-1
Thus 2f (x) ^-1 = x
Or f-1(x) = x/2

G(x)= 4x-3
Let g(x) = x and x = g(x)V-1
Therefore x=4g(X)^-1-3
Rearranging g(x)^-1=(x+3)/4

G(fx)=
G(2x)=4(2x)-3
G(2x)=8x-3
Let g(fx)=x and x = (g(fx))^-1
X=8((g(fx))^-1)-3
X+3 =8(g(fx))^-1)
Therefore (g(fx))^-1 = (x+3)/8

F^-1(g(x)^-1=
F^-1(x+3)/4= ((x+3)/4)/2
F^-1(x+3)/4=(x+3)/8
Thus F^-1(g(x)^-1=(g(fx))^-1

Answer 2

To simplify these problems let remember that for every linear function f(x) = ax + b
1. The function is one to one corresponding soit has an inverse f^-1
2. It is easy to prove that: f^-1*(x) = (b - x)/a

since f(x) = 2x then f^-1(x) = x/2
Since g(x) = 4x - 3 then g^-1(x)= (3+x)/4
f^-1(g^-1(x)) = ((3+x)/4)/2 = (3+x)/8
g(f(x)) = 4(2x) - 3 = 8x -3
g(f(x))^-1 = (3 + x)/8

Answer 3

here very easy ,can be used the composant of the functions f and g f(g(x))=(f*g)(x) with f(x) =2x,g(x) =4x-3 ,and g(f(x)=4(2x)-3=8x-3 ,then the inverse of this is x=(1/8)y+3/8 ,also inverse of f is x = (i/2)y so

Answer 4

f^-1(f(x)) = x
f^-1(x) = x / 2
g^-1(x) = (x + 3) / 4
f^-1(g^-1(x)) = ((x + 3) / 4) / 2 = (x + 3) / 8
g(f(x)) = 4 (2x) - 3 = 8x - 3
g(f(x))^-1 = (x + 3) / 8 = f^-1(g^-1(x))

Answer 5

g(x)=4x-3
Therefore g^_1(x)=1/(4x-3)
f(x) = 2x
Therefore f^-1(x)=1/(2x) and
f^-1{g^-1x)=1/{2(1/4x-3)}
=(4x-3)/2............................i
f(x)^-1=1/2x
g{f(x)^-1} = 4(1/2x) - 3
= (2/x) - 3......................................ii

Answer 6

How 'bout you go ahead and solve that while I go to the beach with the girls.

laters.........

Answer 7

8x-6

Answer 8

holy wat the **** is that man im only in 8th man!