Suppose D(f) = {x| x ∈ R, x ≠ 3} and f(x) =(x+2) /(x-3). What does define(s) inverse (f−1)?
x = (y+2)/(y-3)
xy - 3x = y + 2
y(x-1) = 3x+2
g(x) = f-1(x) = y = (3x+2)/(x-1)
check: f(x) and g(x) are inverses if (f◦g)(x) = (g◦f)(x) = x
g(f) = (3f+2)/(f-1) = (3(x+2)/(x-3)+2)/((x+2)/(x-3) - 1)
= (3x+6+2x-6)/(x+2-x+3)
= 5x/5
= x
f(g) = (g+2)/(g-3) = ((3x+2)/(x-1) + 2)/((3x+2)/(x-1) - 3)
= (3x+2+2x-2)/(3x+2-3x+3)
= 5x/5
= x
Thanks oobleck
To define the inverse of a function, we need to find an expression for f^(-1)(x) such that when we compose f and f^(-1), we get the identity function.
To find the inverse of f(x) = (x + 2)/(x - 3), we can follow these steps:
Step 1: Replace f(x) with y:
y = (x + 2)/(x - 3)
Step 2: Swap x and y:
x = (y + 2)/(y - 3)
Step 3: Solve for y:
Multiply both sides of the equation by (y - 3) to eliminate the denominator:
x(y - 3) = y + 2
Distribute on the left side:
xy - 3x = y + 2
Move all terms involving y to the left side and all terms involving x to the right side:
xy - y = 3x + 2
Factor out y:
y(x - 1) = 3x + 2
Divide both sides by (x - 1):
y = (3x + 2)/(x - 1)
Step 4: Replace y with f^(-1)(x):
f^(-1)(x) = (3x + 2)/(x - 1)
So, the expression defining the inverse function f^(-1)(x) is (3x + 2)/(x - 1).