Given: f(x) = 2x^2+3f(x)=2x2+3 and f(v) = 5v+6f(v)=5v+6
We know that f(x)f(x) is going to have two inverses; this may cause a problem.
Compute f^-1(x)f−1(x) by substituting f^-1(x)f−1(x) for every xx with within f(x)f(x):
f(f^-1(x)) = 2(f^-1(x))^2+3f(f−1(x))=2(f−1(x))2+3
The property of a function and its inverse, f(f^-1(x)) = xf(f−1(x))=x, causes the left side to become xx:
x = 2(f^-1(x))^2+3x=2(f−1(x))2+3
Solve for f^-1(x)f−1(x):
x-3 = 2(f^-1(x))^2x−3=2(f−1(x))2
(x-3)/2 = (f^-1(x))^2x−32=(f−1(x))2
f^-1(x) = sqrt((x-3)/2)f−1(x)=√x−32 and f^-1(x) = -sqrt((x-3)/2)f−1(x)=−√x−32
Substitute x = f(v)x=f(v) into the left sides and x = 5v+6x=5v+6 into the right sides:
f^-1(f(v)) = sqrt((5v+6-3)/2)f−1(f(v))=√5v+6−32 and f^-1(f(v)) = -sqrt((5v+6-3)/2)f−1(f(v))=−√5v+6−32
Use the property f^-1(f(v)) = vf−1(f(v))=v to make the left sides become v:
v = sqrt((5v+6-3)/2)v=√5v+6−32 and v = -sqrt((5v+6-3)/2)v=−√5v+6−32
We are about to square both equations but this will eliminate, therefore, the two equations degenerate into a single equation:
v^2 = (5v+6-3)/2v2=5v+6−32
2v^2= 5v+32v2=5v+3
2v^2- 5v-3=02v2−5v−3=0
Factor:
(2v+1)(v-3) = 0(2v+1)(v−3)=0
v = -1/2v=−12 and v=3v=3
