The interval in which the function f(x) = {x^ | vedclass

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(B) Let $f(x) = {x^2}{e^{ - x}}$.To find the interval where the function is non-decreasing,we find the derivative $f'(x)$ and set $f'(x) \ge 0$.Using

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$[0, 2]$

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(B) Let $f(x) = {x^2}{e^{ - x}}$.To find the interval where the function is non-decreasing,we find the derivative $f'(x)$ and set $f'(x) \ge 0$.Using the product rule: $f'(x) = \frac{d}{dx}({x^2}) \cdot {e^{ - x}} + {x^2} \cdot \frac{d}{dx}({e^{ - x}})$.$f'(x) = 2x{e^{ - x}} - {x^2}{e^{ - x}} = {e^{ - x}}(2x - {x^2}) = x{e^{ - x}}(2 - x)$.For the function to be non-decreasing,$f'(x) \ge 0$.Since ${e^{ - x}} > 0$ for all real $x$,the sign of $f'(x)$ depends on $x(2 - x)$.$x(2 - x) \ge 0 \implies x(x - 2) \le 0$.This inequality holds when $x$ is between the roots $0$ and $2$,inclusive.Thus,the interval is $[0, 2]$.

The interval in which the function $f(x) = {x^2}{e^{ - x}}$ is non-decreasing is:

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