F (x + 1) = the square of X + 2x (x is greater than 0) to find the inverse function of y = f (x + 1)

F (x + 1) = the square of X + 2x (x is greater than 0) to find the inverse function of y = f (x + 1)

y=f(x+1)
= x^2 + 2x
= x^2 + 2x + 1 -1
= (x+1)^2 - 1
So x = √ (y + 1) - 1
That is, the inverse function y = [√ (x + 1)] - 1

Finding the inverse function of F (x) = 4x / X-5

y=4x/(x-5)
(x-5)y=4x -> (y-4)x=5y -> x=5y/(y-4)
So the inverse function is f (x) = 5x / (x-4)

If f (4x + 1) = x + 2, what is the inverse function f ^ - 1 (x) of F

f(4x+1)=x+2
Let 4x + 1 = t
4x=t-1
x=(t-1)/4
f(4x+1)=f(t)=(t-1)/4+2
∴f(x)=(x-1)/4+2
f(x)-2=(x-1)/4
x-1=4[f(x)-2]
x=4f(x)-8+1=4f(x)-7
∴f^-1(x)=4x-7