Problema Solution
The equation defines a one-to-one function f. f(x) = 1/x − 8
Determine f^-1 (x).
Verify that f compose f −1 and f −1 compose f are both the identity function.
(f compose f −1)(x) = f(f −1(x))
(f −1 compose f)(x) = f −1(f(x))
Answer provided by our tutors
Since f is 1-1 function then there is inverse function f^-1 defined by x = f^-1(y) if and only if y = f(x).
Procedure for Determining the Equation for f^-1 from the Equation of f
A. Substitute y for f(x)
y = 1/x - 8
B. Interchange the x and y
x = 1/y - 8
C. Solve, if possible, for y in terms of x
x = 1/y - 8 => 1/y = x + 8 => y = 1/(x + 8)
D. Substitute f^-1(x) for y
f^-1(x) = 1/(x + 8)
E. Verify that Df = Rf^-1 and Rf = Df^-1 (D denotes the domain, R denotes the range of the function)
Df^-1 = R/{-8} = Rf
Rf^-1 = R/{0} = Df
Lets prove that f(f ^−1(x)) and f ^−1(f(x)) are the identity functions
Indeed
for x element of Df^-1
f(f ^−1(x)) = f(1/(x + 8)) = 1/(1/(x + 8)) - 8 = x + 8 - 8 = x
for x element of Df
f ^−1(f(x)) = f^-1(1/x − 8) = 1/(1/x − 8 + 8) = 1/(1/x) = x