a graphing calculator is recommended.\nconsider the following function. (if an answer does not exist, enter…

Explanation:

Step1: Find the first derivative

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{2}), (u^\prime=2x), (v = e^{-x}), (v^\prime=-e^{-x}). (f^\prime(x)=2xe^{-x}-x^{2}e^{-x}=xe^{-x}(2 - x)) Set (f^\prime(x)=0), then (x = 0) or (x = 2) (since (e^{-x}\gt0) for all (x)).

Step2: Analyze the sign of (f^\prime(x))

For (0\lt x\lt2), (f^\prime(x)\gt0), so (f(x)) is increasing. For (x\gt2), (f^\prime(x)\lt0), so (f(x)) is decreasing. At (x = 0), (f(0)=0^{2}e^{-0}=0). At (x = 2), (f(2)=4e^{-2}=\frac{4}{e^{2}}). As (x\to+\infty), (\lim_{x\rightarrow+\infty}x^{2}e^{-x}=\lim_{x\rightarrow+\infty}\frac{x^{2}}{e^{x}}), using L - H rule twice: First application: (\lim_{x\rightarrow+\infty}\frac{2x}{e^{x}}), second application: (\lim_{x\rightarrow+\infty}\frac{2}{e^{x}} = 0). So the minimum value of (f(x)) for (x\geq0) is (0) (at (x = 0)) and the maximum value is (\frac{4}{e^{2}}) (at (x = 2)).

Step3: Find the second derivative

(f^\prime(x)=xe^{-x}(2 - x)=2xe^{-x}-x^{2}e^{-x}) (f^{\prime\prime}(x)=2e^{-x}-2xe^{-x}-2xe^{-x}+x^{2}e^{-x}=e^{-x}(x^{2}-4x + 2)) Set (f^{\prime\prime}(x)=0), then (x^{2}-4x + 2=0). Using the quadratic formula (x=\frac{4\pm\sqrt{16 - 8}}{2}=2\pm\sqrt{2}). We want to find where (f(x)) increases most rapidly (i.e., where (f^\prime(x)) has a maximum). Since (x\geq0), we consider (x = 2-\sqrt{2}) (because the function (y = f^\prime(x)) is a quadratic - like function (after factoring out (e^{-x})) and we take the value in the domain (x\geq0)).

Answer:

(a) Minimum: (0), Maximum: (\frac{4}{e^{2}}) (b) (x = 2-\sqrt{2})