Question:

Step 1: Identify the given functions and what we need to find. We are given: - $f(x) = \ln\left(\frac{1+x}{1-x}\right)$ where $f:(-1,1) \to \mathbb{R}$ - $g(x) = \frac{3x+x^3}{1+3x^2}$ where $g:\mathbb{R} \to (-1,1)$ We need to find $f(g(x))$. Step 2: Apply the function $f$ to $g(x)$. By definition of composition, we substitute $g(x)$ into $f$: $$f(g(x)) = \ln\left(\frac{1+g(x)}{1-g(x)}\right)$$ Step 3: Substitute the expression for $g(x)$. $$f(g(x)) = \ln\left(\frac{1+\frac{3x+x^3}{1+3x^2}}{1-\frac{3x+x^3}{1+3x^2}}\right)$$ Step 4: Simplify the numerator of the argument. The numerator becomes: $$1+\frac{3x+x^3}{1+3x^2} = \frac{1+3x^2+3x+x^3}{1+3x^2}$$ Recognize that the numerator is a perfect cube: $$1+3x^2+3x+x^3 = (1+x)^3$$ Step 5: Simplify the denominator of the argument. The denominator becomes: $$1-\frac{3x+x^3}{1+3x^2} = \frac{1+3x^2-3x-x^3}{1+3x^2}$$ Recognize that this is also a perfect cube: $$1+3x^2-3x-x^3 = (1-x)^3$$ Step 6: Substitute the simplified forms back into the logarithm. $$f(g(x)) = \ln\left(\frac{\frac{(1+x)^3}{1+3x^2}}{\frac{(1-x)^3}{1+3x^2}}\right) = \ln\left(\frac{(1+x)^3}{(1-x)^3}\right)$$ Step 7: Apply logarithm properties to simplify. Using the property $\ln\left(\frac{a^n}{b^n}\right) = n\ln\left(\frac{a}{b}\right)$: $$f(g(x)) = 3\ln\left(\frac{1+x}{1-x}\right) = 3f(x)$$ **Final Answer:** $f(g(x)) = 3f(x)$ This matches **Option 3**.