How to find the inverse of a quadratic equation?

In general, quadratic equations that represent a parabola that opens up or down do not have an inverse because for any given value of y there are two corresponding values of x (except for the vertex). However, the domain restriction #-3 < x < -1# allows us to find an inverse.

#f(x) = -(x+1)^2-1, -3 < x < -1#

Begin by substituting #x = f^-1(x)# everywhere that you see an x:

#f(f^-1(x)) = -(f^-1(x)+1)^2 -1, -3 < f^-1(x) < -1#

The left side becomes x because of the property of all inverses, #f(f^-1(x)) = x#:

#x = -(f^-1(x)+1)^2 -1, -3 < f^-1(x) < -1#

Add 1 to both sides:

#x+ 1 = -(f^-1(x)+1)^2, -3 < f^-1(x) < -1#

Multiply both sides by -1:

#-x-1 = (f^-1(x)+1)^2, -3 < f^-1(x) < -1#

Take the square root of both sides:

#+-sqrt(-x-1) = f^-1(x)+1, -3 < f^-1(x) < -1#

Please observe that this is why, in general, quadratics do not have an inverse. The above is really two equations:

#sqrt(-x-1) = f^-1(x)+1# and #-sqrt(-x-1) = f^-1(x)+1#

But we have a range restriction that tells us that we must pick the negative case:

#-sqrt(-x-1) = f^-1(x)+1, -3 < f^-1(x) < -1#

Flip the equation:

# f^-1(x)+1= -sqrt(-x-1), -3 < f^-1(x) < -1#

Subtract 1 from both sides:

# f^-1(x) = -sqrt(-x-1)-1, -3 < f^-1(x) < -1#