Show that (f of g)⁻¹(x) = (g⁻¹ of f⁻¹)(x) using f(x)=5x-2 and g(x)= 2x+7?

Question

math help

Answer 1

Assuming that you want the inverse functions where the square symbol is, start by finding (f o g)(x) and then computing the inverse function.  Do the same thing for (g^(-1) of f^(-1))(x) and compare the answers.

f(x) = 5x - 2

g(x) = 2x + 7

(f o g)(x) = f(2x + 7) = 5(2x + 7) - 2 = 10x + 35 - 2 = 10x + 33

To find the inverse function, switch x and y and solve for y.

x = 10y + 33

x - 33 = 10y

x/10 - 3.3 = y

(f o g)^(-1)(x) = x/10 - 3.3

Now find both g^(-1)(x) and f^(-1)(x).  Again, first find g(x) and f(x) and switch x and y and solve for y to get the inverse function.

g(x) = 2x + 7

x = 2y + 7

x - 7 = 2y

x/2 - 3.5 = y

g^(-1)(x) = x/2 - 3.5

f(x) = 5x - 2

x = 5y - 2

x + 2 = 5y

x/5 + .4 = y

f^(-1)(x) = x/5 + .4

Since you know g^(-1)(x) and f^(-1)(x) you can compute (g^(-1) of f^(-1))(x) as shown below.

(g^(-1)(f^(-1)(x))) = g^(-1)(x/5 + .4) = (x/5 + .4)/2 - 3.5 = x/10 + .2 - 3.5 = x/10 - 3.3

Then

(f o g)^(-1)(x) = x/10 - 3.3 = (g^(-1)(f^(-1)(x)))

Hope this helps you!