Let f(x) = int_0^{x^2} frac{t^2-8t+15}{e^t} d | vedclass

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(A) Using the Leibniz rule for differentiation under the integral sign,we have:$f'(x) = \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}} \cdot \frac{d}{dx}(x^2)

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$2$ and $3$

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(A) Using the Leibniz rule for differentiation under the integral sign,we have:$f'(x) = \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}} \cdot \frac{d}{dx}(x^2) - \frac{0^2 - 8(0) + 15}{e^0} \cdot \frac{d}{dx}(0)$$f'(x) = \frac{x^4 - 8x^2 + 15}{e^{x^2}} \cdot (2x)$$f'(x) = \frac{(x^2 - 3)(x^2 - 5)(2x)}{e^{x^2}}$$f'(x) = \frac{2x(x - \sqrt{3})(x + \sqrt{3})(x - \sqrt{5})(x + \sqrt{5})}{e^{x^2}}$The critical points are $x = -\sqrt{5}, -\sqrt{3}, 0, \sqrt{3}, \sqrt{5}$.Analyzing the sign of $f'(x)$ across these intervals:For $x For $-\sqrt{5}  0$. (Local Min at $-\sqrt{5}$)For $-\sqrt{3} For $0  0$. (Local Min at $0$)For $\sqrt{3} For $x > \sqrt{5}$,$f'(x) > 0$. (Local Min at $\sqrt{5}$)Thus,there are $2$ local maximum points and $3$ local minimum points.

Let $f(x) = \int_0^{x^2} \frac{t^2-8t+15}{e^t} dt$,$x \in R$. Then the numbers of local maximum and local minimum points of $f$,respectively,are:

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