Question

For $f(x)=x+\dfrac{1}{x}$ $(x\neq0)$

• A. Local maximum value is 2
• B. Local minimum value is −2
• C. Local maximum value is −2
• D. Local minimum value<local maximum value

Hint:

For functions containing \(x+\frac{1}{x}\), stationary points often occur at \(x=\pm1\). Always verify maximum/minimum using the second derivative test.

Answer 1

The Correct Option is C

To determine the local maximum and minimum values of the function \(f(x) = x + \dfrac{1}{x}\) for \(x \neq 0\), we need to follow these steps:

1. Calculate the derivative of the function to find critical points where potential extrema can occur. The derivative of \(f(x)\) is: \(f'(x) = 1 - \dfrac{1}{x^2}\).
2. Set the derivative equal to zero to find critical points: \(1 - \dfrac{1}{x^2} = 0\). This leads to:

\(\dfrac{1}{x^2} = 1\) or \(x^2 = 1\)

1. Thus, \(x = 1\) or \(x = -1\).
2. Evaluate the function at the critical points to find the function values:
   • At \(x = 1\), \(f(1) = 1 + \dfrac{1}{1} = 2\).
   • At \(x = -1\), \(f(-1) = -1 + \dfrac{1}{-1} = -2\).
3. Determine the nature of these critical points (maximum or minimum) using the second derivative test:
   • Calculate the second derivative: \(f''(x) = \dfrac{2}{x^3}\).
   • Evaluate the second derivative at \(x = 1\): \(f''(1) = \dfrac{2}{1^3} = 2 > 0\). Positive value indicates a local minimum at \(x = 1\).
   • Evaluate the second derivative at \(x = -1\): \(f''(-1) = \dfrac{2}{(-1)^3} = -2 < 0\). Negative value indicates a local maximum at \(x = -1\).

Conclusion: The function \(f(x) = x + \dfrac{1}{x}\) has a local maximum value of \(-2\) at \(x = -1\) and a local minimum value of \(2\) at \(x = 1\).

Therefore, the correct answer is: The local maximum value is −2.