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

Step 1: Differentiate the function.
Given \[ f(x)=x+\frac{1}{x} \]
Derivative:
\[ f'(x)=1-\frac{1}{x^2} \]
Step 2: Find critical points.
Set derivative equal to zero:
\[ 1-\frac{1}{x^2}=0 \]
\[ x^2=1 \]
\[ x=\pm1 \]
Step 3: Use second derivative test.
Second derivative:
\[ f''(x)=\frac{2}{x^3} \]
Check at \(x=1\):
\[ f''(1)=2>0 \] Therefore \(x=1\) gives local minimum.
Value:
\[ f(1)=1+1=2 \]
Check at \(x=-1\):
\[ f''(-1)=-2<0 \] Therefore \(x=-1\) gives local maximum.
Value:
\[ f(-1)=-1-1=-2 \]
Step 4: Conclusion.
Local minimum value = \(2\)
Local maximum value = \(-2\)
Hence the correct statement is:
\[ \text{Local maximum value is } -2 \]
Final Answer: $\boxed{-2}$