If f(x) = ln (x + \(\sqrt{1+\text{x}^2}\) ), then which one of the following is correct ?
f(x) + f(−x) = 0
Analyzing the Function Problem
The question asks us to find the correct relationship between \(f(x)\) and \(f(-x)\) for the given function \(f(x) = \ln (x + \sqrt{1+x^2})\). This involves evaluating the function at \(-x\) and comparing the result to the original function \(f(x)\).
Solving for the Function Relationship
Step 1: Evaluate \(f(-x)\)
We substitute \(-x\) into the expression for \(f(x)\):
\[f(-x) = \ln ((-x) + \sqrt{1+(-x)^2})\]Since \( (-x)^2 = x^2 \), this simplifies to:
\[f(-x) = \ln (-x + \sqrt{1+x^2})\]Step 2: Examine the Given Options
The options suggest relationships between \(f(x)\) and \(f(-x)\), such as sums or differences equaling zero, or one being a multiple of the other. We will test these relationships using the expressions for \(f(x)\) and \(f(-x)\).
Step 3: Check Option 1: \(f(x) + f(-x) = 0\)
Let's add \(f(x)\) and \(f(-x)\):
\[f(x) + f(-x) = \ln (x + \sqrt{1+x^2}) + \ln (-x + \sqrt{1+x^2})\]Using the logarithm property \(\ln a + \ln b = \ln (ab)\), we combine the terms:
\[f(x) + f(-x) = \ln \left[ (x + \sqrt{1+x^2}) (-x + \sqrt{1+x^2}) \right]\] \[f(x) + f(-x) = \ln \left[ (\sqrt{1+x^2} + x) (\sqrt{1+x^2} - x) \right]\]This expression is in the form \((a+b)(a-b) = a^2 - b^2\), where \(a = \sqrt{1+x^2}\) and \(b = x\). Applying this identity:
\[f(x) + f(-x) = \ln \left[ (\sqrt{1+x^2})^2 - x^2 \right]\] \[f(x) + f(-x) = \ln \left[ (1+x^2) - x^2 \right]\] \[f(x) + f(-x) = \ln \left[ 1+x^2 - x^2 \right]\] \[f(x) + f(-x) = \ln (1)\]Since \(\ln(1) = 0\), we have:
\[f(x) + f(-x) = 0\]This confirms that the relationship in Option 1 is correct.
Checking Other Options
Since \(f(x) + f(-x) = 0\), it means \(f(-x) = -f(x)\).
- Option 2: \(f(x) - f(-x) = 0\) implies \(f(x) = f(-x)\). This is not true, as \(f(-x) = -f(x)\).
- Option 3: \(2f(x) = f(-x)\) implies \(2f(x) = -f(x)\), which means \(3f(x) = 0\) or \(f(x) = 0\). This is not true for all \(x\) in the domain of \(f(x)\).
- Option 4: \(f(x) = 2f(-x)\) implies \(f(x) = 2(-f(x))\), which means \(f(x) = -2f(x)\), or \(3f(x) = 0\) or \(f(x) = 0\). This is not true for all \(x\).
Therefore, the only correct relationship is \(f(x) + f(-x) = 0\).
Understanding Function Symmetry
A function \(f(x)\) is called an odd function if \(f(-x) = -f(x)\) for all \(x\) in its domain. Our finding that \(f(x) + f(-x) = 0\) is equivalent to \(f(-x) = -f(x)\). Thus, the function \(f(x) = \ln (x + \sqrt{1+x^2})\) is an odd function.
A function \(f(x)\) is called an even function if \(f(-x) = f(x)\) for all \(x\) in its domain.
Revision Table: Key Concepts in Function Properties
| Concept | Definition | Example Property |
|---|---|---|
| Odd Function | \(f(-x) = -f(x)\) for all \(x\) | Graph is symmetric about the origin |
| Even Function | \(f(-x) = f(x)\) for all \(x\) | Graph is symmetric about the y-axis |
| Logarithm Property | \(\ln a + \ln b = \ln (ab)\) | Used to combine log terms |
| Difference of Squares | \((a+b)(a-b) = a^2 - b^2\) | Used to simplify algebraic expressions |
Additional Information: Properties of Logarithms and Functions
The natural logarithm function, \(\ln(y)\), is defined for \(y > 0\). For \(f(x) = \ln (x + \sqrt{1+x^2})\), the argument \(x + \sqrt{1+x^2}\) must be greater than 0. We know that \(\sqrt{1+x^2} > \sqrt{x^2} = |x|\). So, \(\sqrt{1+x^2} > -x\). This implies \(x + \sqrt{1+x^2} > 0\) for all real \(x\). Thus, the domain of \(f(x)\) is all real numbers, \((-\infty, \infty)\).
The fact that \(f(x)\) is an odd function means its graph has rotational symmetry about the origin. If a point \((a, b)\) is on the graph, then the point \( (-a, -b) \) is also on the graph.
Similar Questions
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