Suppose that the inverse of the function \(f(x) = \sqrt{ax + b}\) is given by \(f^{-1}(x) = -(a+2)x^2+(2b+1)\) for all \(x\)satisfying \(ax+b \ge 0\). Find \(a+b\).
Much Help would be appreciated!
y = sqrt (ax + b) square both sides
y^2 = ax + b subtract b from both sides
y^2 - b = ax divide both sides by a
y^2 / a - b/a = x "swap" x and y
x^2 / a - b/a = the inverse
So.....we have that
(1/a)x^2 - b/a = -(a + 2)x^2 + (2b + 1)
This must mean that
1/a = -(a + 2)
1/a = -a - 2
a + 1/a + 2 = 0 multiply through by a and rearrange
a^2 + 2a + 1 = 0 factor
(a + 1)^2 = 0 take the root
a + 1 = 0
a = -1
So.....we also have that
-b/a = 2b + 1
-b/ (-1) = 2b + 1
b = 2b + 1
-1 = b
So
a + b =
-1 + -1 =
-2