Question

If \( f(x) = \frac{x}{\sqrt{1 + x^2}} \), then \( (f \circ f)(x) \) is:

• A. \( \frac{3x}{1 + x^2} \)
• B. \( \frac{x}{\sqrt{1 + 3x^2}} \)
• C. \( \frac{3x}{\sqrt{1 - x^2}} \)
• D. None of these

Hint:

When finding the composition of functions, substitute the output of the first function into the second function and simplify.

Answer 1

The Correct Option is B

Given the function: \[ f(x) = \frac{x}{\sqrt{1 + x^2}} \] We need to find \( (f \circ f)(x) \), which means substituting \( f(x) \) into itself. That is: \[ (f \circ f)(x) = f(f(x)) \] Substitute \( f(x) = \frac{x}{\sqrt{1 + x^2}} \) into the function \( f(x) \): \[ f(f(x)) = f\left( \frac{x}{\sqrt{1 + x^2}} \right) \] Now, substitute \( \frac{x}{\sqrt{1 + x^2}} \) into the expression for \( f(x) \): \[ f\left( \frac{x}{\sqrt{1 + x^2}} \right) = \frac{\frac{x}{\sqrt{1 + x^2}}}{\sqrt{1 + \left( \frac{x}{\sqrt{1 + x^2}} \right)^2}} \] Simplifying the denominator: \[ \left( \frac{x}{\sqrt{1 + x^2}} \right)^2 = \frac{x^2}{1 + x^2} \] Thus, the denominator becomes: \[ \sqrt{1 + \frac{x^2}{1 + x^2}} = \sqrt{\frac{1 + x^2 + x^2}{1 + x^2}} = \sqrt{\frac{1 + 3x^2}{1 + x^2}} \] So, the function \( f(f(x)) \) simplifies to: \[ f(f(x)) = \frac{x}{\sqrt{1 + 3x^2}} \] Thus, the correct answer is \( \frac{x}{\sqrt{1 + 3x^2}} \).