Key Concepts and Formulas

• Function Evaluation: To find $f(g(x))$, substitute $g(x)$ for every instance of $x$ in the definition of $f(x)$.
• Logarithm Properties: The key property used here is the power rule: $\log_b(A^k) = k \log_b(A)$.
• Algebraic Identities: Perfect square trinomials: $(a-b)^2 = a^2 - 2ab + b^2$ and $(a+b)^2 = a^2 + 2ab + b^2$.

Step-by-Step Solution

Step 1: Substitute the argument into the function definition. We are given $f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$ and need to find $f\left( {{{2x} \over {1 + {x^2}}}} \right)$. We substitute $\frac{2x}{1+x^2}$ for $x$ in the definition of $f(x)$: $f\left( {{{2x} \over {1 + {x^2}}}} \right) = {\log _e}\left( {{{1 - \left( {{{2x} \over {1 + {x^2}}}} \right)} \over {1 + \left( {{{2x} \over {1 + {x^2}}}} \right)}}} \right)$ Explanation: This step directly applies the definition of function evaluation. The expression $\frac{2x}{1+x^2}$ becomes the new input for the function $f$.

Step 2: Simplify the argument of the logarithm. We focus on simplifying the fraction inside the logarithm: $\frac{1 - \frac{2x}{1+x^2}}{1 + \frac{2x}{1+x^2}}$ To simplify this complex fraction, we find a common denominator for the numerator and the denominator separately.

• Numerator simplification: $1 - \frac{2x}{1+x^2} = \frac{1(1+x^2)}{1+x^2} - \frac{2x}{1+x^2} = \frac{1+x^2 - 2x}{1+x^2} = \frac{x^2 - 2x + 1}{1+x^2}$
• Denominator simplification: $1 + \frac{2x}{1+x^2} = \frac{1(1+x^2)}{1+x^2} + \frac{2x}{1+x^2} = \frac{1+x^2 + 2x}{1+x^2} = \frac{x^2 + 2x + 1}{1+x^2}$ Now, we divide the simplified numerator by the simplified denominator: $\frac{\frac{x^2 - 2x + 1}{1+x^2}}{\frac{x^2 + 2x + 1}{1+x^2}} = \frac{x^2 - 2x + 1}{1+x^2} \times \frac{1+x^2}{x^2 + 2x + 1} = \frac{x^2 - 2x + 1}{x^2 + 2x + 1}$ Explanation: We used the principle of finding a common denominator to combine terms and then performed fraction division. The terms $(1+x^2)$ in the numerator and denominator cancel out.

Step 3: Apply algebraic identities to further simplify the argument. We recognize that the numerator and denominator are perfect square trinomials:

• $x^2 - 2x + 1 = (x-1)^2 = (1-x)^2$
• $x^2 + 2x + 1 = (x+1)^2$ So, the argument simplifies to: $\frac{(1-x)^2}{(1+x)^2} = \left( \frac{1-x}{1+x} \right)^2$ Explanation: This step leverages the standard algebraic identities for $(a-b)^2$ and $(a+b)^2$ to rewrite the expression in a more compact form.

Step 4: Substitute the simplified argument back into the logarithm and use logarithm properties. Now, we substitute this simplified expression back into the logarithm from Step 1: $f\left( {{{2x} \over {1 + {x^2}}}} \right) = {\log _e}\left( {\left( {{{1 - x} \over {1 + x}}} \right)^2} \right)$ Using the logarithm power rule, $\log_b(A^k) = k \log_b(A)$: $= 2 {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$ Explanation: The power rule for logarithms allows us to bring the exponent of the argument to the front as a multiplier.

Step 5: Relate the result to the original function $f(x)$. We observe that the expression ${\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$ is the definition of $f(x)$. Therefore, we can rewrite the result as: $2 {\log _e}\left( {{{1 - x} \over {1 + x}}} \right) = 2f(x)$ Explanation: By comparing the simplified expression with the original definition of $f(x)$, we establish the relationship.

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and signs. A small error can lead to a completely different result.
• Logarithm Properties Misapplication: Ensure that the power rule is applied correctly. It applies to the entire argument being raised to a power, not to terms within the argument or a power of the logarithm itself (e.g., $(\log_e x)^2$).
• Domain Considerations: The condition $|x| < 1$ ensures that $1-x > 0$ and $1+x > 0$, so $\frac{1-x}{1+x} > 0$, making the logarithm well-defined. It also ensures that $\left| \frac{2x}{1+x^2} \right| < 1$, so the argument of $f$ remains within its valid domain.

Summary

The problem required us to evaluate $f\left( {{{2x} \over {1 + {x^2}}}} \right)$ given $f(x) = {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$. By substituting the new argument into the function, simplifying the resulting complex fraction using algebraic identities, and applying the power rule of logarithms, we arrived at $2 {\log _e}\left( {{{1 - x} \over {1 + x}}} \right)$, which is equal to $2f(x)$.

The final answer is \boxed{2f(x)} which corresponds to option (B).