Key Concepts and Formulas

• First Derivative Test: To find local extrema, find critical points (where $f'(x) = 0$ or is undefined) and analyze the sign changes of $f'(x)$ around these points.
• Fundamental Theorem of Calculus (Part 1) & Chain Rule: If $f(x) = \int_a^{g(x)} h(t) \, dt$, then $f'(x) = h(g(x)) \cdot g'(x)$.
• Local Maxima and Minima: If $f'(x)$ changes from positive to negative at $x=c$, then $f(x)$ has a local maximum at $x=c$. If $f'(x)$ changes from negative to positive at $x=c$, then $f(x)$ has a local minimum at $x=c$.

Step-by-Step Solution

Step 1: Find the First Derivative, $f'(x)$

We need to find the derivative of $f(x)$ to locate potential local extrema. We use the Fundamental Theorem of Calculus and the Chain Rule since $f(x)$ is defined as an integral with a variable upper limit. $f(x) = \int_0^{x^2} \frac{t^2 - 8t + 15}{e^t} \, dt$ Let $g(x) = x^2$ and $h(t) = \frac{t^2 - 8t + 15}{e^t}$. Then $g'(x) = 2x$. Applying the formula $f'(x) = h(g(x)) \cdot g'(x)$, we have: $f'(x) = \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}} \cdot (2x) = \frac{2x(x^4 - 8x^2 + 15)}{e^{x^2}}$ Thus, $f'(x) = \frac{2x(x^4 - 8x^2 + 15)}{e^{x^2}}$

Step 2: Identify Critical Points

Critical points are where $f'(x) = 0$ or is undefined. Since $e^{x^2}$ is always positive and never zero, $f'(x)$ is defined for all $x$. We need to find where $f'(x) = 0$. $\frac{2x(x^4 - 8x^2 + 15)}{e^{x^2}} = 0$ This implies $2x(x^4 - 8x^2 + 15) = 0$. We can factor the quartic polynomial by letting $y = x^2$: $2x(y^2 - 8y + 15) = 0$ $2x(y - 3)(y - 5) = 0$ Substituting $x^2$ back in for $y$: $2x(x^2 - 3)(x^2 - 5) = 0$ The critical points are: $2x = 0 \implies x = 0$ $x^2 - 3 = 0 \implies x = \pm\sqrt{3}$ $x^2 - 5 = 0 \implies x = \pm\sqrt{5}$ So the critical points are $x = -\sqrt{5}, -\sqrt{3}, 0, \sqrt{3}, \sqrt{5}$.

Step 3: Analyze the Sign of $f'(x)$ (First Derivative Test)

We will now determine the sign of $f'(x)$ in the intervals defined by the critical points to determine where $f(x)$ is increasing or decreasing. Since $e^{x^2}$ is always positive, the sign of $f'(x)$ depends only on the sign of $2x(x^2 - 3)(x^2 - 5)$.

• $x < -\sqrt{5}$: Let $x = -3$. Then $2x < 0$, $x^2 - 3 > 0$, and $x^2 - 5 > 0$. So $f'(x) < 0$.
• $-\sqrt{5} < x < -\sqrt{3}$: Let $x = -2$. Then $2x < 0$, $x^2 - 3 > 0$, and $x^2 - 5 < 0$. So $f'(x) > 0$.
• $-\sqrt{3} < x < 0$: Let $x = -1$. Then $2x < 0$, $x^2 - 3 < 0$, and $x^2 - 5 < 0$. So $f'(x) < 0$.
• $0 < x < \sqrt{3}$: Let $x = 1$. Then $2x > 0$, $x^2 - 3 < 0$, and $x^2 - 5 < 0$. So $f'(x) > 0$.
• $\sqrt{3} < x < \sqrt{5}$: Let $x = 2$. Then $2x > 0$, $x^2 - 3 > 0$, and $x^2 - 5 < 0$. So $f'(x) < 0$.
• $x > \sqrt{5}$: Let $x = 3$. Then $2x > 0$, $x^2 - 3 > 0$, and $x^2 - 5 > 0$. So $f'(x) > 0$.

Step 4: Classify Local Extrema

Using the First Derivative Test:

• $x = -\sqrt{5}$: $f'(x)$ changes from negative to positive $\implies$ local minimum.
• $x = -\sqrt{3}$: $f'(x)$ changes from positive to negative $\implies$ local maximum.
• $x = 0$: $f'(x)$ changes from negative to positive $\implies$ local minimum.
• $x = \sqrt{3}$: $f'(x)$ changes from positive to negative $\implies$ local maximum.
• $x = \sqrt{5}$: $f'(x)$ changes from negative to positive $\implies$ local minimum.

Step 5: Conclusion

We have 2 local maximum points ($x = -\sqrt{3}$ and $x = \sqrt{3}$) and 3 local minimum points ($x = -\sqrt{5}$, $x = 0$, and $x = \sqrt{5}$).

Common Mistakes & Tips

• Missing the $2x$ factor: Don't forget the $2x$ term when finding the derivative and critical points. It's a common mistake to only consider the quartic factor.
• Sign Errors: Be careful when evaluating the sign of $f'(x)$ in each interval, especially with negative values of $x$.
• Factoring the Quartic: Recognize that $x^4 - 8x^2 + 15$ is a quadratic in $x^2$, making it easier to factor.

Summary

We found the derivative of the given integral using the Fundamental Theorem of Calculus and the Chain Rule. Then, we identified the critical points by setting the derivative equal to zero. Finally, we used the First Derivative Test to analyze the sign changes of the derivative around the critical points, allowing us to classify them as local maxima or minima. We found 2 local maxima and 3 local minima.

The final answer is \boxed{2 and 3}, which corresponds to option (A).