Given: f(x)=2x2+3 and f(v)=5v+6
We know that f(x) is going to have two inverses; this may cause a problem.
Compute f−1(x) by substituting f−1(x) for every x with within f(x):
f(f−1(x))=2(f−1(x))2+3
The property of a function and its inverse, f(f−1(x))=x, causes the left side to become x:
x=2(f−1(x))2+3
Solve for f−1(x):
x−3=2(f−1(x))2
x−32=(f−1(x))2
f−1(x)=√x−32 and f−1(x)=−√x−32
Substitute x=f(v) into the left sides and x=5v+6 into the right sides:
f−1(f(v))=√5v+6−32 and f−1(f(v))=−√5v+6−32
Use the property f−1(f(v))=v to make the left sides become v:
v=√5v+6−32 and v=−√5v+6−32
We are about to square both equations but this will eliminate, therefore, the two equations degenerate into a single equation:
v2=5v+6−32
2v2=5v+3
2v2−5v−3=0
Factor:
(2v+1)(v−3)=0
v=−12 and v=3
