Let $f(x) = \displaystyle\int_0^{x^2}\frac{t^2-8t+15}{e^t}\,dt$, $x\in\mathbb{R}$. Then the numbers of local maximum and local minimum points of $f$, respectively, are:
Step-by-Step Solution
Key Concept: By the chain rule $f'(x) = \dfrac{(x^2)^2-8x^2+15}{e^{x^2}}\cdot 2x = \dfrac{2x(x^2-3)(x^2-5)}{e^{x^2}}$; apply the wavy-curve method at $x = 0, \pm\sqrt{3}, \pm\sqrt{5}$ to count sign changes.
$f'(x) = \dfrac{x^4-8x^2+15}{e^{x^2}}\cdot 2x = \dfrac{2x(x^2-5)(x^2-3)}{e^{x^2}}.$
Zeros at $x = -\sqrt{5},-\sqrt{3},0,\sqrt{3},\sqrt{5}$.
Wavy-curve on $(-\infty,\infty)$ (positive leading coefficient, odd degree in numerator):
$f'>0$ on $(-\infty,-\sqrt{5})$, $f'<0$ on $(-\sqrt{5},-\sqrt{3})$, $f'>0$ on $(-\sqrt{3},0)$, $f'<0$ on $(0,\sqrt{3})$, $f'>0$ on $(\sqrt{3},\sqrt{5})$, $f'<0$ on $(\sqrt{5},\infty)$.
Local maxima at $x = -\sqrt{5}, \sqrt{3}$ (2 points); local minima at $x = -\sqrt{3}, 0, \sqrt{5}$ (3 points).
Correct Answer: 1