First of all, I will assume you meant:
F^-1(x) = (1-5x)/(2x), or else I would have just reduced to F^-1(x) = 1 - 5/2 = -3/2 , a constant
The inverse of the inverse of a function would be the function, that is,
the inverse of F^-1(x) would be F(x)
so we just need the inverse of y = (1-5x)/(2x)
interchange the x and y's
x = (1-5y)/(2y)
2xy = 1 - 5y
2xy + 5y = 1
y(2x + 5) = 1
y = 1/(2x+5)
F(x) = 1/(2x + 5)
check: let x = 5
F^-1(5) = (1 - 25)/10 = -2.4
F(-2.4) = 1/(-4.8 + 5) = 1/(.2) = 5
It is highly likely that my answer is correct
find an expression for F(x) and state the domain of F
when F^-1(x)= 1-5x/2x for 0 < x <= 2
2 answers
Just saw the domain part ... 0 < x ≤ 2
F^-1(0) = undefined, but F^-1(x) ----> negative infinity
F^-1(2) = (1-10)/4 = -9/4
so the domain of F(x) is
- infinity < x < -9/4
F^-1(0) = undefined, but F^-1(x) ----> negative infinity
F^-1(2) = (1-10)/4 = -9/4
so the domain of F(x) is
- infinity < x < -9/4