Question

The minimum value of \(f(x)=x+\dfrac{4}{x+2}\) is

• A. \(-1\)
• B. \(-2\)
• C. \(1\)
• D. \(2\)

Hint:

To find maxima or minima of a function, first find critical points using \(f'(x)=0\), then apply the second derivative test.

Answer 1

The Correct Option is D

Step 1: Write the given function.
The function is \[ f(x)=x+\frac{4}{x+2} \] We need to find its minimum value.

Step 2: Differentiate the function.
Differentiate with respect to \(x\): \[ f'(x)=1-\frac{4}{(x+2)^2} \] For extrema, \[ f'(x)=0 \] Hence, \[ 1-\frac{4}{(x+2)^2}=0 \] \[ \frac{4}{(x+2)^2}=1 \] \[ (x+2)^2=4 \] \[ x+2=\pm2 \] Thus, \[ x=0 \quad \text{or} \quad x=-4 \]

Step 3: Use second derivative test.
Differentiate again: \[ f''(x)=\frac{8}{(x+2)^3} \] At \(x=0\), \[ f''(0)=\frac{8}{8}=1\gt 0 \] Hence, \(x=0\) gives a minimum value.
At \(x=-4\), \[ f''(-4)=\frac{8}{(-2)^3}=-1\lt 0 \] Hence, \(x=-4\) gives a maximum value.

Step 4: Find the minimum value.
Substitute \(x=0\) into the function: \[ f(0)=0+\frac{4}{2} \] \[ =2 \]

Step 5: Final Answer.
Therefore, the minimum value is \[ \boxed{2} \]