The one-to-one function f is defined below.

Question

f(x)= x/8x-3

Find f^-1(x),where f^-1 is the inverse of f.

Also state the domain and range of f^-1 in interval notation.

f^-1(x)=

Domain of f^-1=

Range of f^-1=

Answer 1

f(x) = x / (8x-3)

Domain f = (-∞, 3/8) ∪ (3/8, ∞) = Range f^-1

y = f(x) has a horizontal asymptote at y = 1/8

Range f^-1 = (-∞, 1/8) ∪ (1/8, ∞) = Domain f^-1

To find f^-1(x), let y = x / (8x-3)

Switch x and y: x = y/(8y-3)

Solve for y: 8xy - 3x = y

8xy - y = 3x

y(8x - 1) = 3x

y = 3x / (8x - 1)

f^-1(x) = 3x / (8x - 1)