Given the function f(x) = 2x^2 +3f(x)=2x2+3. Find values of v such that f(v)= 5v + 6f(v)=5v+6?

1 Answer
Jun 26, 2018

v = -1/2v=12 and v=3v=3

Explanation:

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)f1(x) by substituting f^-1(x)f1(x) for every xx with within f(x)f(x):

f(f^-1(x)) = 2(f^-1(x))^2+3f(f1(x))=2(f1(x))2+3

The property of a function and its inverse, f(f^-1(x)) = xf(f1(x))=x, causes the left side to become xx:

x = 2(f^-1(x))^2+3x=2(f1(x))2+3

Solve for f^-1(x)f1(x):

x-3 = 2(f^-1(x))^2x3=2(f1(x))2

(x-3)/2 = (f^-1(x))^2x32=(f1(x))2

f^-1(x) = sqrt((x-3)/2)f1(x)=x32 and f^-1(x) = -sqrt((x-3)/2)f1(x)=x32

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)f1(f(v))=5v+632 and f^-1(f(v)) = -sqrt((5v+6-3)/2)f1(f(v))=5v+632

Use the property f^-1(f(v)) = vf1(f(v))=v to make the left sides become v:

v = sqrt((5v+6-3)/2)v=5v+632 and v = -sqrt((5v+6-3)/2)v=5v+632

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+632

2v^2= 5v+32v2=5v+3

2v^2- 5v-3=02v25v3=0

Factor:

(2v+1)(v-3) = 0(2v+1)(v3)=0

v = -1/2v=12 and v=3v=3