Let \( f(x) = \int_0^{x^2 \frac{t^2 - 8t + 15}{e^t}} dt, \, x \in \mathbb{R} \). Then the numbers of local maximum and local minimum points of \( f \), respectively, are:
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For problems involving perpendicular distance from a point to a line in three-dimensional space, use the formula:
\[
d = \frac{| \mathbf{a} \cdot (\mathbf{r_0} - \mathbf{r_1}) |}{|\mathbf{a}|}
\]
Where:
- \( \mathbf{a} \) is the direction vector of the line,
- \( \mathbf{r_0} \) is the point,
- \( \mathbf{r_1} \) is any point on the line.
This will help you calculate the distance effectively.
To determine the number of local maxima and minima of the function \( f(x) \), we need to first find its first and second derivatives. We start by differentiating \( f(x) \) using the Leibniz rule for the derivative of an integral: \[ f'(x) = \frac{d}{dx} \left( \int_0^{x^2} \frac{t^2 - 8t + 15}{e^t} dt \right) = \frac{d}{dx} \left( \int_0^{x^2} g(t) dt \right), \] where \( g(t) = \frac{t^2 - 8t + 15}{e^t} \). By the Leibniz rule, this becomes: \[ f'(x) = 2x \cdot \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}}. \] Now, for critical points, we solve \( f'(x) = 0 \): \[ 2x \cdot \frac{x^4 - 8x^2 + 15}{e^{x^2}} = 0. \] This equation will hold if either \( x = 0 \) or \( x^4 - 8x^2 + 15 = 0 \). Solving the quadratic equation \( x^4 - 8x^2 + 15 = 0 \) leads to: \[ x^2 = \frac{8 \pm \sqrt{64 - 60}}{2} = 4 \pm \sqrt{1}, \] which gives solutions \( x^2 = 5 \) and \( x^2 = 3 \), or \( x = \pm \sqrt{5}, \pm \sqrt{3} \).
Now, we proceed to check the nature of these critical points using the second derivative \( f''(x) \) to determine whether they correspond to local maxima or minima. After calculating \( f''(x) \) and analyzing its sign at the critical points \( x = 0, \pm \sqrt{3}, \pm \sqrt{5} \), we find that there are 2 local maxima and 2 local minima.
Thus, the number of local maxima and minima is 2 and 2, respectively.
