The function $f(x)$ satisfies
\[3f(x) - 2 f \left( \frac{1}{x} \right) = x\]for all $x \neq 0.$  Find $f(4).$
Setting $x = 4,$ we get
\[3f(4) - 2 f \left( \frac{1}{4} \right) = 4.\]Setting $x = \frac{1}{4},$ we get
\[3 f \left( \frac{1}{4} \right) - 2f(4) = \frac{1}{4}.\]We can view these equations as a system in $f(4)$ and $f \left( \frac{1}{4} \right).$  Solving this system, we find $f(4) = \boxed{\frac{5}{2}}.$
\frac{5}{2}
Level 4
Intermediate Algebra
Problem: The function $f(x)$ satisfies
\[3f(x) - 2 f \left( \frac{1}{x} \right) = x\]for all $x \neq 0.$  Find $f(4).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the function \( f(x) \), we start with the given equation \( 3f(x) - 2f\left(\frac{1}{x}\right) = x \). We can derive additional equations by substituting different values for \( x \).

First, let’s substitute \( x = 1 \):
\[
3f(1) - 2f(1) = 1 \implies f(1) = 1.
\]

Next, substitute \( x = 4 \):
\[
3f(4) - 2f\left(\frac{1}{4}\right) = 4.
\]

Now, substitute \( x = \frac{1}{4} \):
\[
3f\left(\frac{1}{4}\right) - 2f(4) = \frac{1}{4}.
\]

We now have a system of two equations:
1. \( 3f(4) - 2f\left(\frac{1}{4}\right) = 4 \) (Equation 1)
2. \( 3f\left(\frac{1}{4}\right) - 2f(4) = \frac{1}{4} \) (Equation 2)

From Equation 1, we can express \( f\left(\frac{1}{4}\right) \) in terms of \( f(4) \):
\[
2f\left(\frac{1}{4}\right) = 3f(4) - 4 \implies f\left(\frac{1}{4}\right) = \frac{3f(4) - 4}{2}.
\]

Substituting this expression into Equation 2 gives:
\[
3\left(\frac{3f(4) - 4}{2}\right) - 2f(4) = \frac{1}{4}.
\]
Multiplying through by 2 to eliminate the fraction:
\[
3(3f(4) - 4) - 4f(4) = \frac{1}{2}.
\]
Expanding and simplifying:
\[
9f(4) - 12 - 4f(4) = \frac{1}{2} \implies 5f(4) - 12 = \frac{1}{2}.
\]
Adding 12 to both sides:
\[
5f(4) = 12 + \frac{1}{2} = \frac{24}{2} + \frac{1}{2} = \frac{25}{2}.
\]
Dividing by 5:
\[
f(4) = \frac{25}{10} = \frac{5}{2}.
\]

Thus, the final answer is \( \frac{5}{2} \).<|END_OF_LATENT|> The final answer is $\frac{5}{2}$.