Step 1: Analyze graph behavior near origin. The graph passes through the origin, which suggests that $f(0) = 0$. Step 2: Analyze behavior for positive $x$. For $x>0$, the function rises to a positive peak and then decays towards zero. This matches the form $x . 2^{-x}$, where the exponential decay dominates for large $x$. Step 3: Analyze behavior for negative $x$. For $x<0$, the graph goes below the axis (negative values) and approaches $0$ as $x \to -\infty$. This matches $x . 2^{-|x|}$, because for $x<0$, $-|x| = x$, giving $f(x) = x . 2^{x}$, which is negative but approaches 0 as $x \to -\infty$. Step 4: Eliminate other options. - (A) $x^{2} 2^{-|x|}$ is always non-negative (since $x^{2} \geq 0$), but the graph shows negative values for $x<0$. Wrong. - (C) $|x| 2^{-x}$ is always non-negative as well. Wrong. - (D) $x 2^{-x}$ is not symmetric with respect to $x<0$, and doesn’t match the decay behavior. Wrong. Hence, the correct function is (B). Final Answer: \[ \boxed{f(x) = x 2^{-|x|}} \]
