Let $f(x)=\int_{0}^{x^{2}}\frac{t^{2}-8t+15}{e^{t}}dt$, $x \in R$. Then the number of local maximum and local minimum points of $f$, respectively, are:
- (1) 2 and 3
- (2) 3 and 2
- (3) 1 and 3
- (4) 2 and 2
Solution:
Using Leibniz rule to differentiate $f(x)$:
$$ f'(x) = \frac{(x^2)^2 – 8(x^2) + 15}{e^{x^2}} \cdot \frac{d}{dx}(x^2) – 0 $$
$$ f'(x) = \frac{x^4 – 8x^2 + 15}{e^{x^2}} \cdot 2x $$
$$ f'(x) = \frac{(x^2-3)(x^2-5) \cdot 2x}{e^{x^2}} $$
Critical points are where $f'(x) = 0$:
$$ 2x(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{5})(x+\sqrt{5}) = 0 $$
Points: $x = -\sqrt{5}, -\sqrt{3}, 0, \sqrt{3}, \sqrt{5}$.
Sign scheme of $f'(x)$:
– For $x < -\sqrt{5}$: $f'(x) < 0$ (5 negative terms) – $(-\sqrt{5}, -\sqrt{3})$: $f'(x) > 0$
– $(-\sqrt{3}, 0)$: $f'(x) < 0$ – $(0, \sqrt{3})$: $f'(x) > 0$
– $(\sqrt{3}, \sqrt{5})$: $f'(x) < 0$ – $x > \sqrt{5}$: $f'(x) > 0$
Minima (change from – to +): $x = -\sqrt{5}, 0, \sqrt{5}$ (3 points).
Maxima (change from + to -): $x = -\sqrt{3}, \sqrt{3}$ (2 points).
Total: 2 Maxima, 3 Minima.
Ans. (1)