Problem 50 - A Textbook of Differential Calculus

Question

Statement I The range of \(\log \left(\frac{1}{1+x^{2}}\right)\) is \((-\infty, \infty)\). Statement II When \(0

Step-by-Step Solution

Answer

Statement I is false; Statement II is true.

Step 1: Understanding the function

Let's first understand the function given in Statement I: \( f(x) = \log \left(\frac{1}{1+x^2}\right) \). This involves taking the logarithm of a fraction, where the numerator is 1 and the denominator is \(1 + x^2\). Since \(x^2 \geq 0\), \(1 + x^2 \geq 1\). Hence, \(\frac{1}{1+x^2} \leq 1\), meaning that \( f(x) \leq 0 \). The function cannot take values greater than 0 and never reaches \(-\infty\).

Step 2: Determine the actual range

To find the range of \( f(x) = \log \left(\frac{1}{1+x^2}\right) \), consider its components: as \( x^2 \to \infty \), \( \frac{1}{1+x^2} \to 0\), so \( \log \frac{1}{1+x^2} \to -\infty \). As \( x \to 0 \), \( \frac{1}{1+x^2} = 1 \), so \( \log 1 = 0 \). Therefore, the range of \( f(x) \) is \((-\infty, 0]\).

Step 3: Analyze Statement II

Statement II claims that when \(0 < x \leq 1\), \( \log x \in (-\infty, 0] \). Since the logarithm of a number less than 1 is negative and the logarithm of 1 is 0, this statement is true. For \( 0 < x \leq 1\), \(\log x < 0\) or \(\log 1 = 0\), thus covering the range \((-infty, 0]\).

Step 4: Conclude the correctness of the statements

Now we verify the correctness: Statement I is incorrect as its range does not cover \((-infty, \infty)\). Statement II is correct because it correctly describes the behavior of \(\log x\) for \(0 < x \leq 1\).

Key Concepts

Logarithmic Functions

Logarithmic functions are important tools in differential calculus because they transform multiplicative processes into additive ones. This property is especially useful for simplifying complex equations that involve multiplication. A general logarithmic function is of the form \( y = \log_b(x) \), where \( b \) is the base of the logarithm. The most commonly used bases are 10 (common logarithm) and \( e \) (natural logarithm). Logarithmic functions have some interesting characteristics:

• They only take positive numbers as inputs, because you can't take the logarithm of a zero or negative number in real numbers.
• The output of a logarithmic function can be any real number.
• Logarithms transform exponential growth or decay into a straight line, making them easier to analyze.

In our exercise, the function \( f(x) = \log \left(\frac{1}{1+x^2}\right) \) uses the natural logarithm. It comprises understanding both the logarithmic nature and the fraction \( \frac{1}{1+x^2} \), where \( x^2 \geq 0 \). This tells us the fraction is always less than or equal to one, making the logarithm expression always non-positive.

Function Range

The range of a function is the set of all possible output values. For the logarithmic function in our exercise, \( f(x) = \log \left(\frac{1}{1+x^2}\right) \), the range is determined by the behavior of the fraction \( \frac{1}{1+x^2} \). Since \( x^2 \geq 0 \), \( 1 + x^2 \geq 1 \), and thus \( \frac{1}{1+x^2} \leq 1 \).

As \( x \) approaches 0, the value of the fraction approaches 1, leading the logarithm to take its maximum value of \( \log 1 = 0 \). Conversely, as \( x^2 \) becomes very large, \( \frac{1}{1+x^2} \) approaches 0, causing the logarithm to infinitely decrease towards \(-\infty\). Therefore, the range of the function is \((-\infty, 0]\). This contradicts Statement I that suggests the range is \((-\infty, \infty)\), highlighting an important analysis in understanding function behavior.

Mathematical Proof

Mathematical proof is the process of demonstrating that a certain statement or proposition is true using logical reasoning. In calculus, proofs help validate claims about functions, their behavior, and their properties. When dissecting exercises, such as our statements, we employ logical steps to prove or disprove claims.

In our example, we looked at the behavior of \( f(x) = \log \left(\frac{1}{1+x^2}\right) \) to understand its range. By considering different limits \( (x \to 0 \text{ and } x^2 \to \infty) \), we proved the range is \((-\infty, 0]\). This logical sequence involved verifying the possible outcomes and limits of the fraction in the logarithm, providing a comprehensive analysis of why Statement I was false and Statement II was true.

Inequality Analysis

Inequality analysis is key in understanding and verifying mathematical statements, particularly those involved in calculus. Inequalities help compare the size or order of two values. In our problem, understanding inequalities allowed us to accurately define the behavior of logarithmic functions.

For Statement II, exploration of \( \log x \) where \( 0 < x \leq 1 \) was valid because \( \log x \) is negative for any \( x < 1 \), and 0 when \( x = 1 \). By analyzing and applying inequalities, we verified the claim that \( \log x \in (-\infty, 0] \).

Using inequalities to analyze \( f(x) \) is crucial because it assists in understanding possible value developments and transitions, solidifying comprehension of range, limits, and mathematical truths.