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Question 1209954: Suppose f(x) is a rational function such that
3 f \left( \frac{1}{x} \right) - \frac{f(x)}{x} = x
for all $x \neq 0$. Find f(-2).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the given equation be
$$3 f \left( \frac{1}{x} \right) - \frac{f(x)}{x} = x \quad (*)$$
for all $x \neq 0$.
Replace $x$ with $\frac{1}{x}$ in (*):
$$3 f(x) - x f \left( \frac{1}{x} \right) = \frac{1}{x} \quad (**)$$
From (*), we have
$$3 f \left( \frac{1}{x} \right) = x + \frac{f(x)}{x}$$
$$f \left( \frac{1}{x} \right) = \frac{x}{3} + \frac{f(x)}{3x}$$
Substitute this into (**):
$$3 f(x) - x \left( \frac{x}{3} + \frac{f(x)}{3x} \right) = \frac{1}{x}$$
$$3 f(x) - \frac{x^2}{3} - \frac{f(x)}{3} = \frac{1}{x}$$
$$9 f(x) - x^2 - f(x) = \frac{3}{x}$$
$$8 f(x) = x^2 + \frac{3}{x}$$
$$f(x) = \frac{x^2}{8} + \frac{3}{8x}$$
Now, we need to find $f(-2)$:
$$f(-2) = \frac{(-2)^2}{8} + \frac{3}{8(-2)}$$
$$f(-2) = \frac{4}{8} - \frac{3}{16}$$
$$f(-2) = \frac{1}{2} - \frac{3}{16}$$
$$f(-2) = \frac{8}{16} - \frac{3}{16}$$
$$f(-2) = \frac{5}{16}$$
Therefore, $f(-2) = \frac{5}{16}$.
Final Answer: The final answer is $\boxed{\frac{5}{16}}$
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