Answer :
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.
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.
