求使 $y = 3x^2 e^{5-2x}$ 为递减函数的 $x$ 的取值范围。

1. Differentiate the function with respect to $x$ using the product rule and chain rule

$\frac{dy}{dx} = (3x^2)\frac{d}{dx}(e^{5-2x}) + (e^{5-2x})\frac{d}{dx}(3x^2)$
$\frac{dy}{dx} = 3x^2(-2e^{5-2x}) + e^{5-2x}(6x)$

2. Factorise the derivative

$\frac{dy}{dx} = -6x^2e^{5-2x} + 6xe^{5-2x}$
$\frac{dy}{dx} = 6xe^{5-2x}(1 - x)$

3. Set the derivative to be less than zero for a decreasing function and solve the inequality

For a decreasing function, $\frac{dy}{dx} < 0$.
$6xe^{5-2x}(1 - x) < 0$
Since $e^{5-2x} > 0$ for all real $x$, we have:
$x(1 - x) < 0$
$x(x - 1) > 0$
$x < 0$ or $x > 1$