Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\frac{x^{2}-1}{x^{2}+1}\), for every real number \(x\), then the minimum value of \(f\) (A) does not exist because \(f\) is unboundecd (B) is not attained even though \(f\) is bounded (C) is equal to 1 (D) is equal to \(-1\)

Step-by-Step Solution

Answer

The minimum value of \( f(x) \) is \(-1\).

Step 1: Analyze the Function

Given function is \( f(x) = \frac{x^{2} - 1}{x^{2} + 1} \). We need to find its minimum value by first analyzing its behavior.

Step 2: Consider the Range

The expression \( \frac{x^2 - 1}{x^2 + 1} \) must be examined by considering the limits as \( x \to \pm \infty \) and when \( x = 0 \). At extreme \( x \) values, the function's behavior approximates to 1 since \( x^2 - 1 \approx x^2 + 1 \) results in \( 1 \).

Step 3: Derivative Approach for Critical Points

To find critical points, differentiate \( f(x) \) and set \( f'(x) = 0 \). Suppose \( f'(x) = \frac{(2x)(x^2+1) - (x^2-1)(2x)}{(x^2+1)^2} \). Simplifies to \( \frac{2x}{(x^2+1)^2} (x^2+1 - x^2 +1) = \frac{4x}{(x^2+1)^2} = 0 \). This implies \( x = 0 \) is a critical point.

Step 4: Evaluate Critical Point

Evaluate \( f(x) \) at \( x = 0 \): \( f(0) = \frac{0^2 - 1}{0^2 + 1} = -1 \).

Step 5: Evaluate the Function Behavior

The function \( f(x) \) approaches 1 as \( x \to \pm \infty \). Therefore, \( f(x) \) is bounded and reaches its minimum value at \( x = 0 \).

Step 6: Conclude the Minimum Value

Since the function attains \( -1 \) and \( f(x) \to 1 \) as \( x \to \pm \infty \), the minimum value is \( -1 \), which occurs at \( x = 0 \).

Key Concepts

Critical Points

Critical points of a function are places where the function's rate of change slows down and becomes zero or undefined. They are found using the first derivative. If the derivative of a function is zero (\( f'(x) = 0 \)), the function might have a local maximum, minimum, or a saddle point here.

• First, take the derivative of the function.
• Set the derivative equal to zero to find critical points.
• In this function, \( f(x) = \frac{x^2 - 1}{x^2 + 1} \), setting \( f'(x) = 0 \) yields \( x = 0 \) as a critical point.

Critical points help to determine where these key features occur on the graph of the function. It's crucial to check what happens around these points to understand the function's behavior fully.

Derivative

The derivative is a mathematical tool that gives information about the rate of change of a function. It helps analyze how a function increases or decreases and can identify critical points.

• For the given function, \( f(x) = \frac{x^2 - 1}{x^2 + 1} \), the derivative is found using the quotient rule.
• The derivative \( f'(x) = \frac{4x}{(x^2+1)^2} \) is calculated by differentiating the numerator and denominator and applying the quotient rule.
• Setting this derivative equal to zero helps find potential minimums and maximums.

Derivatives tell us about the slope of the function's graph. At critical points, this slope is zero, meaning the graph flattens out, which is useful for finding extremum values – the maximum or minimum points.

Function Behavior

Understanding a function's behavior involves analyzing how the function acts as the independent variable changes. This includes identifying trends as the variable approaches infinity and zero.

• For \( f(x) = \frac{x^2 - 1}{x^2 + 1} \), as \( x \to \pm \infty \), the function approaches 1.
• Near the critical point \( x = 0 \), the function drops to a minimum value.

Observing the behavior at \( x = 0 \) and as \( x \to \pm \infty \) reveals that the function is bounded. This means the values of \( f(x) \) are restricted to a certain range, and won't go beyond specific values. These observations are vital for understanding how the function behaves across different intervals.

Range of a Function

The range of a function is the set of all possible output values it can produce. For our function \( f(x) \), determining the range helps identify possible minimum or maximum values.

• For the function \( f(x) = \frac{x^2 - 1}{x^2 + 1} \), we find the output range by considering behavior at critical points and as \( x \to \pm \infty \).
• As shown, the function has a minimum value of \( -1 \) at \( x = 0 \).
• As \( x \) approaches infinity or negative infinity, \( f(x) \) approaches 1.

This range highlights that the function is bounded between -1 and 1. Knowing the range is crucial for fully understanding how a function operates and what outcomes it can produce. It helps us predict the limits beyond which the function does not exist.