Answer: Yes.
If the inverses of two functions are both functions, then the inverse of the sum or difference of the original functions will also be a function.
Explanation:
Let
f(x) is the first function, and its inverse is f⁻¹(x), so that f(f⁻¹)(x) = x.
Similarly,
g(x) s the second function, and its inverse s g⁻¹(x), so that g(g⁻¹)(x) = x.
Then
(a) If h(x) = f(x) + g(x), then both h(x) and h⁻¹(x) are functions.
(b) If h(x) = f(x) - g(x), then both h(x) and h⁻¹(x) are functions.
Example.
Let f(x) = 3x + 5 and g(x) = 2x - 7.
It can be shown easily that
f⁻¹(x) = (x-5)/3, and g⁻¹(x) = (x+7)/2.
Let h(x) = f(x)+g(x) = 5x - 2
Create the inverse.
y = 5x - 2
Set x = 5y - 2 and solve for y.
h⁻¹(x) = y = (x+2)/5
Test h(h⁻¹)(x).
x=1: h(1) = 5*1 - 2 = 3
h⁻¹(1) = (1+2)/5 = 3/5
h(h⁻¹)(1) = 5*(3/5) - 2 = 1 Correct
x=-2; h(-2) = 5*(-2) - 2 = -12
h⁻¹(-2) = (-2 + 2)/5 =0
h(h⁻¹)(-2) = 5*0 - 2 = -2 Correct
Similarly, by setting h(x) = f(x) - g(x), it can be shown that h(h⁻¹)(x) = x.
A graph showing the result is given below.
Clearly, both f⁻¹(x) + g⁻¹(x), and f⁻¹(x) - g⁻¹(x) pass the vertical line test which is required for functions.
