How to find the inverse of a quadratic equation?

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

1 Answer
Douglas K.
Jun 15, 2018

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:

±√−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:

√−x−1=f−1(x)+1 and −√−x−1=f−1(x)+1

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

−√−x−1=f−1(x)+1,−3<f−1(x)<−1

Flip the equation:

f−1(x)+1=−√−x−1,−3<f−1(x)<−1

Subtract 1 from both sides:

f−1(x)=−√−x−1−1,−3<f−1(x)<−1

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