We are given the functional equation:
\[ f(x) + 2f\left(\frac{1}{x}\right) = 3x \]
We are asked to find the sum of all possible values of $x$ for which $f(x) = 3$.
Substitute $f(x) = 3$ into the equation:
\[ 3 + 2f\left(\frac{1}{x}\right) = 3x \]
Solve for $f\left(\frac{1}{x}\right)$:
\[ 2f\left(\frac{1}{x}\right) = 3x - 3 \]
\[ f\left(\frac{1}{x}\right) = \frac{3x - 3}{2} \]
Now, substitute $x = \frac{1}{x}$ into the original equation:
\[ f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x} \]
This results in a system of equations, which can be solved to find the value of $x$. After solving the system, we find that the sum of all possible values of $x$ for which $f(x) = 3$ is -3.
Let \( f \) be an injective map with domain x, y, z and range 1, 2, 3 such that exactly one of the following statements is correct and the remaining are false.}
[I.] \( f(x) = 1 \Rightarrow f(y) = 1, f(z) = 2 \)
[II.] \( f(x) = 2 \Rightarrow f(y) = 1, f(z) = 1 \)
[III.] \( f(x) = 1, f(y) = 1, f(z) = 2 \) Then the value of \( f(1) \) is:
The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} x & \text{for } 0 \le x \le 1 \\ 1 & \text{for } x \ge 1 \\ 0 & \text{otherwise} \end{cases} \]
The properties of the function are as follows:
\[ f_1(x) = f(-x) \quad \text{for all } x \] \[ f_2(x) = -f(x) \quad \text{for all } x \] \[ f_3(x) = f(f(x)) \quad \text{for all } x \]