can anyone help me to solve this additional maths problems??

Question

the questions is :

given the function h : x = 2x -3/x+1, where x ≠ 1.
find the function of h^-1 (x).

please help me!!

urgh!! i need some help! its make my head gonna blow

Answer 1

h(x) = (2x - 3)/(x + 1) // multiply by (x + 1)

(x + 1)h(x)=(2x - 3)

x*h(x) + h(x) = 2x - 3 // - 2x

x*h(x) - 2x + h(x) = -3

x*(h(x) - 2) + h(x) = -3 // - h(x)

x*(h(x)-2) = -3 - h(x) // divide by h(x)-2

x = (-3 - h(x))/(h(x) - 2)

Or h^-1(x) = (-3 - x)/(x - 2)

Answer 2

given h=2x-(3/x)+1, then the inverse can be obtained by writing x is versus h therefore:

multiplying both sides by x and collecting terms

2 x^2 +(1-h) x -3 =0

or x= -(1-h)/4 +/- (1/4) square root of ( (1-h)^2 +24)

h is a dummy argument one may define two functions by replacing h by x. for

y= -(1-x)/4 +/- (1/4) square root of ( (1-x)^2 +24))

note that for any given h there are two x (except at x=0 which is undefined), i.e., the inverse is not uniquely defined