Problem 169 - A Complete Resource Book in Mathematics for JEE Main

Question

The function \(f(x)=\frac{x}{2}+\frac{2}{x}\) has a local minimum at (A) \(x=2\) (B) \(x=-2\) (C) \(x=0\) (D) \(x=1\)

Step-by-Step Solution

Answer

(A) \(x=2\)

Step 1: Find the derivative of the function

To find the local minimum, start by finding the first derivative of the function \(f(x) = \frac{x}{2} + \frac{2}{x}\). Use the power rule and the derivative of \( \frac{1}{x} \). The derivative is: \[ f'(x) = \frac{1}{2} - \frac{2}{x^2} \]

Step 2: Solve for critical points

Set the first derivative equal to zero to find the critical points:\[ \frac{1}{2} - \frac{2}{x^2} = 0 \]Solve for \( x \):\[ \frac{1}{2} = \frac{2}{x^2} \]\[ x^2 = 4 \] \[ x = \pm2 \]

Step 3: Determine whether the critical points are local minima or maxima

Use the second derivative test to determine the nature of the critical points. Find the second derivative of \( f(x) \).\[ f''(x) = \frac{4}{x^3} \]Evaluate the second derivative at the critical points:For \( x = 2 \):\[ f''(2) = \frac{4}{8} = \frac{1}{2} > 0 \] (Local minimum)For \( x = -2 \):\[ f''(-2) = -\frac{1}{2} < 0 \] (Not a minimum)

Key Concepts

Understanding the Derivative

The derivative is a fundamental concept in calculus that highlights how a function changes. To find the derivative of a function like \(f(x)=\frac{x}{2}+\frac{2}{x}\), we use the power rule. This involves finding the rate of change of the function at any point \(x\) and involves differentiating each term separately.
For example, the derivative of \(\frac{x}{2}\) is simply \(\frac{1}{2}\), as the derivative of \(x\) is 1 and multiplying it by \(\frac{1}{2}\) gives this result. For \(\frac{2}{x}\) or \(2x^{-1}\), using the power rule gives us a derivative of \(-2x^{-2}\) or \(-\frac{2}{x^2}\).
Combining these, the derivative of \(f(x)\) is \(f'(x) = \frac{1}{2} - \frac{2}{x^2}\).

Understanding how derivatives work helps us analyze how functions behave at different points and is crucial for finding critical points.

Identifying Critical Points

Critical points of a function occur where its derivative equals zero or is undefined. These points are important because they can indicate local maxima, minima, or points of inflection.
In the function \(f(x) = \frac{x}{2} + \frac{2}{x}\), the critical points are found by solving \(f'(x) = 0\).

• This leads to the equation \(\frac{1}{2} - \frac{2}{x^2} = 0\).
• Simplifying gives us \(x^2 = 4\), resulting in \(x = \pm2\).

Thus, the critical points are \(x = 2\) and \(x = -2\). We can use these points to further investigate the nature of the function at these positions using additional tests.

The Second Derivative Test

The second derivative test provides insight into the concavity of the function at a critical point. It helps to determine whether these points are local minima, maxima, or points of inflection. To use this test, calculate the second derivative \(f''(x)\).
For \(f(x) = \frac{x}{2} + \frac{2}{x}\), we already have the first derivative as \(f'(x) = \frac{1}{2} - \frac{2}{x^2}\). The second derivative is \(f''(x) = \frac{4}{x^3}\). Evaluate \(f''(x)\) at each critical point:

• For \(x = 2\), \(f''(2) = \frac{4}{8} = \frac{1}{2} > 0\), indicating \(x = 2\) is a local minimum since the function is concave up.
• For \(x = -2\), \(f''(-2) = -\frac{1}{2} < 0\), indicating \(x = -2\) is not a minimum, as it is concave down here.

Using the second derivative test presents a clear understanding of the function's behavior at its critical points, assisting in identifying optimal and turning points.