Question

The one-to-one function h is defined by
h(x)=-8x-7/-5-3x
Find h^-1 , the inverse of h . Then, give the domain and range of h^-1 using interval notation.

Answer 1

f(x) = (9x+8) / ( -9+8x)

y = (9x+8) / ( -9+8x)

to find the inverse 1. replace xby y and yby x
2. solve for y
3. replace y by f^-1(x)

1. x=( 9y+8) / (-9+8y)

2. -9x +8xy = 9y +8

8xy-9y = 8+9x

y(8x-9) = (8+9x)

y = (8+9x) / (8x-9)

3. f^-1(x) = (8+9x) / (8x-9)

domain : set of all real values except
x=9/8)
range: set of all real values except 9/8 ( f(x) cannot =9/8)
because in this graph there is a vertical asymptotes at x=9/8and horizontal asymptotes at y=9/8

Answer 2

@pratu043 thanx

Answer 3

h^-1 (x)=?
domain (h^-1)=?
range (h^1)=?