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:

When identifying graphs:

Check symmetry (even/odd functions).
Verify behavior near \(x=0\).
Analyze growth/decay for large \(|x|\).
These help quickly eliminate wrong options.

Answer 1

The Correct Option is A

Step 1: Observe the symmetry of the graph. The function shown is symmetric about the \(y\)-axis, meaning it is an even function. Therefore, the function must involve \(|x|\) or \(x^2\).
Step 2: Eliminate odd functions. Option (B) \(f(x) = x 2^{-|x|}\) and (D) \(f(x) = x 2^{-x}\) are odd functions, not symmetric about the \(y\)-axis. So, these are not correct.
Step 3: Check behavior near \(x=0\). The graph touches the \(x\)-axis at \(x=0\). For \(f(x) = x^2 2^{-|x|}\), we have \(f(0) = 0^2 \cdot 2^0 = 0\), matching the graph.
Step 4: Growth and decay. For large \(|x|\), the exponential factor \(2^{-|x|}\) decays rapidly, and the function tends to \(0\). This matches the graph where values go to \(0\) as \(x \to \pm \infty\). The quadratic term \(x^2\) causes the "bell-like" rise near the origin, before decay sets in.
Thus, the correct representation is \(f(x) = x^2 2^{-|x|}\).