Problem 123 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the range of the function \(f(x)=\frac{x+1}{x^{2}+x+1}\).

Step-by-Step Solution

Answer

The range of the function \(f(x) = \frac{x+1}{x^2+x+1}\) is \(R = [0, \frac{2}{3}]\).

Step 1: Find the domain of the function

The domain of the function consists of all values of \(x\) for which the function is defined. For a rational function, the only values of \(x\) that are not in the domain are those which make the denominator equal to zero. Therefore, we need to solve the equation \(x^2 + x + 1 = 0\) for \(x\). Using the quadratic formula with \(a = 1\), \(b = 1\), and \(c = 1\), we find that: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)}\] Since the discriminant is negative (\(1^2 - 4(1)(1) = -3 < 0\)), the equation has no real roots. Thus, the function is defined for all real numbers and the domain is \(D = (-\infty, \infty)\).

Step 2: Analyze the function's behavior at infinity

As \(x\) approaches infinity (\(x \to \infty\)) or negative infinity (\(x \to -\infty\)), the dominant term in the denominator is \(x^2\), leading to the function approaching 0: \[\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x + 1}{x^2 + x + 1} = 0\] \[\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x + 1}{x^2 + x + 1} = 0\]

Step 3: Find critical points and analyze the function's behavior around them

To find the critical points, we need to find the first derivative of the function and set it equal to zero. Then, we need to analyze the behavior of the function around those critical points. The first derivative is: \[f'(x) = \frac{d}{dx}\left( \frac{x+1}{x^2+x+1}\right)\] Using the quotient rule: \[f'(x) = \frac{(x^2 + x + 1)(1) - (x + 1)(2x + 1)}{(x^2 + x + 1)^2}\] \[f'(x) = \frac{-x^2 + 1}{(x^2 + x + 1)^2}\] Setting \(f'(x) = 0\), we get: \[-x^2 + 1 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1\] Now, we need to analyze the behavior of the function around these critical points. For \(x < -1\): \[f'(x) < 0 \Rightarrow f(x)\] is decreasing. For \(-1 < x < 1\): \[f'(x) > 0 \Rightarrow f(x)\] is increasing. For \(x > 1\): \[f'(x) < 0 \Rightarrow f(x)\] is decreasing.

Step 4: Determine the range of the function

Given the behavior of the function \(f(x)\) on its domain, we know that the function has a minimum value at \(x = -1\) and a maximum value at \(x = 1\). We need to find these values to determine the range: \[f(-1) = \frac{-1 + 1}{(-1)^2 - 1 + 1} = 0\] \[f(1) = \frac{1 + 1}{1^2 + 1 + 1} = \frac{2}{3}\] Since the function is decreasing between \(-\infty\) and \(-1\), then increasing from \(-1\) to \(1\), and then decreasing again from \(1\) to \(\infty\), the range of the function is all real numbers from \(0\) to \(\frac{2}{3}\), including both endpoints. Thus, the range is \(R = [0, \frac{2}{3}]\).

Key Concepts

Domain of a Function

Understanding the domain of a function is essential because it tells us the set of possible input values (usually represented by \( x \)) for which the function is defined. In the context of rational functions, like \( f(x)=\frac{x+1}{x^{2}+x+1} \), the primary concern is ensuring that the denominator doesn't become zero, as this would make the function undefined.

For the function \( f(x) \), we have the quadratic denominator \( x^2 + x + 1 \). To find out where this might be zero, solve \( x^2 + x + 1 = 0 \). Using the quadratic formula:

• \( a = 1 \), \( b = 1 \), and \( c = 1 \)
• Discriminant \( b^2 - 4ac = 1 - 4 \times 1 \times 1 = -3 \)

A negative discriminant indicates no real roots, meaning our denominator never becomes zero for real \( x \). Thus, the domain is all real numbers, or \( (-\infty, \infty) \).

Critical Points

Critical points of a function are where it changes from increasing to decreasing or vice versa. They are found by setting the function's first derivative equal to zero. For \( f(x)=\frac{x+1}{x^{2}+x+1} \), the derivative using the quotient rule is \( f'(x) = \frac{-x^2 + 1}{(x^2 + x + 1)^2} \).

To find the critical points, solve \( -x^2 + 1 = 0 \), which simplifies to \( x = \pm 1 \).

• For \( x = -1 \), the function changes from decreasing to increasing.
• For \( x = 1 \), the function changes from increasing to decreasing.

By evaluating the function at these points, we determine the local minima and maxima within the interval of its domain.

Behavior of Rational Functions

Rational functions, such as \( f(x)=\frac{x+1}{x^{2}+x+1} \), possess unique properties that affect their graphs. Their behavior is influenced by the polynomial expressions in the numerator and denominator. Generally, the degree of polynomials will dictate how the function behaves.

In this function:

• The numerator \( x + 1 \) is linear.
• The denominator \( x^2 + x + 1 \) is quadratic.

As a result, the function tends to zero as \( x \to \pm \infty \). The critical points determined from the first derivative tell us how the function behaves locally. The function decreases from \( -\infty \) to \(-1\), increases between \(-1\) and \(1\), and then decreases again for \( x > 1 \).

These characteristics are crucial for understanding the overall flow and understanding how it approaches its range.

Limits at Infinity

The limits at infinity of a function describe its behavior as \( x \) becomes very large or very small. For rational functions, this often involves examining how the highest degree terms in the numerator and denominator contribute as \( x \to \infty \) or \( x \to -\infty \).

For \( f(x)=\frac{x+1}{x^{2}+x+1} \):

• As \( x \to \infty \), \( f(x) \to 0 \).
• As \( x \to -\infty \), \( f(x) \to 0 \).

The analysis involves recognizing that \( x^2 \), the highest power term in the denominator, grows much faster than any terms in the numerator. This means the "growth" of the denominator outpaces that of the numerator, causing the function value to approach zero. Understanding these limits provides insight into the end-behavior of the function, ensuring a comprehensive grasp of its range and asymptotic behavior.