Question

If h(y)=1+y/2-y, then what is the value of h^-1(5)? Express your answer in simpilest form.

The first step would be to find the inverse of the 1+y/2-y which would be 2y-xy I think, and then would we plug in 5 for x or y?

Answer 1

No, you can't have h^-1 with both x and y in it.

In fact, there are no x's here at all. Just interchange h and y and solve for the new h:

y = (1+h)/(2-h)
y(2-h) = (1+h)
2y - yh = 1+h
h(1+y) = 2y-1
h = (2y-1)/(y+1)

So, h^-1(y) = (2y-1)/(y+1)
h^-1(5) = 9/6 = 3/2

check:
h(3/2) = (5/2)/(1/2) = 5
correct