Question:

Step 1: Analyze the functional equation $f(x) = f(2x)$. We are given that $f(x) = f(2x)$ holds for all $x \in \mathbb{R}$. This means the function value at any point equals its value at double that point. Step 2: Derive an equivalent form by substitution. Replace $x$ with $\frac{x}{2}$ in the functional equation: $$f\left(\frac{x}{2}\right) = f\left(2 \cdot \frac{x}{2}\right) = f(x)$$ This shows that $f\left(\frac{x}{2}\right) = f(x)$ for all $x \in \mathbb{R}$. Step 3: Determine that $f$ must be constant using continuity. From Steps 1 and 2, we have: - $f(x) = f(2x)$ - $f(x) = f\left(\frac{x}{2}\right)$ By repeatedly applying these relations, for any $x \in \mathbb{R}$ and any positive integer $n$: $$f(x) = f(2^n x) = f\left(\frac{x}{2^n}\right)$$ The set $\{2^n x : n \in \mathbb{Z}\}$ forms a dense subset of $\mathbb{R}$ for any $x \neq 0$. Since $f$ is continuous and takes the same value on this dense set, $f$ must be constant everywhere on $\mathbb{R}$. Step 4: Use the given condition to find the constant value. Since $f$ is constant and $f(1) = 3$, we have: $$f(x) = 3 \text{ for all } x \in \mathbb{R}$$ Step 5: Evaluate $f(f(x))$. Since $f(x) = 3$ for all $x$: $$f(f(x)) = f(3) = 3$$ Step 6: Compute the definite integral. $$\int_{-1}^{1} f(f(x))\, dx = \int_{-1}^{1} 3\, dx = 3 \cdot [x]_{-1}^{1} = 3 \cdot (1 - (-1)) = 3 \cdot 2 = 6$$ **Final Answer:** The value of $\displaystyle\int_{-1}^{1} f(f(x))\, dx = 6$ This corresponds to **Option 3: 6**