Question 17.

We are given, $f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$

Substituting $\frac{1}{x}\ for\ x$

$f\left(\dfrac{1}{x}\right)+2f\left(x\right)=\dfrac{3}{x}$

Multiplying the second equation by 2 we will have 

$2f\left(\dfrac{1}{x}\right)+4f\left(x\right)=\dfrac{6}{x}$

Subtracting the first equation from the second equation we have, 

$3f\left(x\right)=\frac{6}{x}-3x$

$f\left(x\right)=\frac{2}{x}-x$

We want the sum of values when this function equals 3, 

$\frac{2}{x}-x=3$

$x^2+3x-2=0$

Since the discriminant is greater than zero, both values of x will be real, and we can directly take the sum of values of $x$ to be, 

$-\frac{3}{1}$

Answer is -3.