Given the function $f(x)=x^{2} e^{-2 x}$ , then at x = 1 the graph of f has
A)a local maximum
B)a local minimum
C)none of these
D)an inflection point

Question

Quizplus · Accepted Answer

DTo determine the nature of the graph at $x = 1$, we first find the first and second derivatives of $f(x) = x^2e^{-2x}$.First derivative:$$f'(x) = \frac{d}{dx}(x^2e^{-2x}) = 2xe^{-2x} + x^2(-2e^{-2x}) = 2xe^{-2x} - 2x^2e^{-2x} = 2xe^{-2x}(1 - x)$$Second derivative:$$f''(x) = \frac{d}{dx}(2xe^{-2x}(1 - x)) = 2e^{-2x}(1 - x) + 2xe^{-2x}(-1) - 4xe^{-2x}(1 - x) = 2e^{-2x}(1 - x - 1 + 2x) = 2e^{-2x}(x)$$At $x = 1$:$$f'(1) = 2(1)e^{-2}(1 - 1) = 0$$This indicates a potential local maximum or minimum.$$f''(1) = 2e^{-2}(1) = 2e^{-2}$$Since $f''(1) > 0$, this indicates that the graph of $f$ has a local minimum at $x = 1$. However, my initial interpretation was incorrect regarding the nature of the point at $x = 1$. Given the positive second derivative, it actually indicates concavity, but the first derivative being zero with a positive second derivative indeed points towards a local minimum, not a maximum or inflection point. The correct interpretation should focus on the signs of the derivatives for accurate classification.

Given the Function f(x)=x2e−2xf(x)=x^{2} e^{-2 x}f(x)=x2e−2x , Then at X = 1 the Graph of F

Given the Function $f(x)=x^{2} e^{-2 x}$ , Then at X = 1 the Graph of F