Statement Study the function f(x)=x e−x/2f(x)=x\,e^{-x/2}f(x)=xe−x/2: find the stationary point (with its nature) and the inflection point. Solution f′(x)=e−x/2+x⋅(−12)e−x/2=e−x/2(1−x2)f'(x)=e^{-x/2}+x\cdot\big(-\tfrac12\big)e^{-x/2}=e^{-x/2}\big(1-\tfrac{x}{2}\big)f′(x)=e−x/2+x⋅(−21)e−x/2=e−x/2(1−2x). It vanishes at x=2x=2x=2. f′′(x)=e−x/2(x4−1)f''(x)=e^{-x/2}\big(\tfrac{x}{4}-1\big)f′′(x)=e−x/2(4x−1); at x=2x=2x=2: f′′(2)=e−1(12−1)<0f''(2)=e^{-1}\big(\tfrac12-1\big)<0f′′(2)=e−1(21−1)<0, hence a maximum. f(2)=2e−1≈0.736f(2)=2e^{-1}\approx 0{.}736f(2)=2e−1≈0.736. Inflection: f′′(x)=0⇒x=4f''(x)=0\Rightarrow x=4f′′(x)=0⇒x=4, f(4)=4e−2≈0.541f(4)=4e^{-2}\approx 0{.}541f(4)=4e−2≈0.541. max (2, 2e−1)≈(2,0.736); inflection (4, 4e−2)≈(4,0.541)\boxed{\text{max }(2,\ 2e^{-1})\approx(2,0{.}736);\ \text{inflection }(4,\ 4e^{-2})\approx(4,0{.}541)}max (2, 2e−1)≈(2,0.736); inflection (4, 4e−2)≈(4,0.541)