Question

How to find the inverse of a quadratic equation?

This is the question: f(x) = −(x + 1)^2 − 1 for −3 < x < −1

Answer 1

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`