Question

Which one of the following options represents the given graph? 

• A. $f(x)=x^2\,2^{-|x|}$
• B. $f(x)=x\,2^{-|x|}$
• C. $f(x)=|x|\,2^{-x}$
• D. $f(x)=x\,2^{-x}$

Hint:

To identify the correct function from a graph: always check symmetry, asymptotic behavior at $\pm \infty$, and positions of maxima/minima. These eliminate wrong options quickly.

Answer 1

The Correct Option is B

Step 1: Symmetry check. The graph is odd, since it is symmetric with a sign change across the origin ($f(-x)=-f(x)$). This immediately eliminates (A) and (C), as both always yield non-negative values.

Step 2: Behavior for $x>0$. For option (B): $f(x)=x\,2^{-x}$ for $x>0$. As $x\to\infty$, $2^{-x}\to0$, so $f(x)\to0^+$. There is a positive maximum near $x=1/\ln2\approx1.44$, consistent with the positive hump in the graph.

Step 3: Behavior for $x<0$. For option (B): $f(x)=x\,2^{x}$ for $x<0$. As $x\to-\infty$, $2^{x}\to0$, hence $f(x)\to0^-$. There is a negative minimum near $x=-1/\ln2\approx-1.44$, consistent with the graph's left-side dip.

Step 4: Eliminate (D). Option (D), $f(x)=x\,2^{-x}$, works fine for $x>0$ but for $x<0$, it diverges to $-\infty$ instead of tending to $0^-$, which does not match the graph. \[ \boxed{\text{Hence the correct function is (B) only.}} \]