Question

If \(f:\mathbb{R}-\{0\}\rightarrow\mathbb{R}\) is a differentiable function such that \[ \frac{1}{3}f(x)+3f\!\left(\frac1x\right) = x-\frac{10}{3}, \] then find \[ f'(3)-f'\!\left(\frac13\right). \]

• A. \(\dfrac{12}{5}\)
• B. \(\dfrac{80}{9}\)
• C. \(3\)
• D. \(5\)

Hint:

For equations involving both \(f(x)\) and \(f(1/x)\), always replace \(x\) by \(1/x\) to obtain a second equation. Solving the resulting system usually leads directly to an explicit formula for \(f(x)\).

Answer 1

The Correct Option is C

Concept: The functional equation contains both \(f(x)\) and \(f(1/x)\). Such problems are usually solved by:
• Replacing \(x\) by \(1/x\).
• Forming simultaneous equations.
• Differentiating the resulting relation.

Step 1: Write the given equation. \[ \frac13 f(x)+3f\!\left(\frac1x\right) = x-\frac{10}{3}. \] Multiplying by \(3\), \[ f(x)+9f\!\left(\frac1x\right) = 3x-10. \] \[ \boxed{f(x)+9f(1/x)=3x-10} \]

Step 2: Replace \(x\) by \(\frac1x\). Then \[ f\!\left(\frac1x\right)+9f(x) = \frac3x-10. \] \[ \boxed{9f(x)+f(1/x)=\frac3x-10} \]

Step 3: Solve the simultaneous equations. Multiply the second equation by \(9\): \[ 81f(x)+9f(1/x) = \frac{27}{x}-90. \] Subtract the first equation: \[ 80f(x) = \frac{27}{x}-90-(3x-10). \] \[ 80f(x) = \frac{27}{x}-3x-80. \] Hence, \[ f(x) = \frac{27/x-3x-80}{80}. \]

Step 4: Differentiate. \[ f'(x) = \frac1{80} \left( -\frac{27}{x^2}-3 \right). \] \[ f'(x) = -\frac{27}{80x^2} -\frac3{80}. \]

Step 5: Evaluate \(f'(3)\). \[ f'(3) = -\frac{27}{80(9)} -\frac3{80} = -\frac3{80} -\frac3{80} = -\frac6{80} = -\frac3{40}. \]

Step 6: Evaluate \(f'\!\left(\frac13\right)\). \[ f'\!\left(\frac13\right) = -\frac{27}{80(1/9)} -\frac3{80} = -\frac{243}{80} -\frac3{80} = -\frac{246}{80}. \]

Step 7: Find the required value. \[ f'(3)-f'\!\left(\frac13\right) = -\frac3{40} +\frac{246}{80}. \] \[ = -\frac6{80} +\frac{246}{80} = \frac{240}{80} = 3. \] Therefore, \[ \boxed{3}. \]