Question

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:

• A. 3 and 2
• B. 2 and 3
• C. 1 and 3
• D. 2 and 2

Hint:

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.

Answer 1

The Correct Option is D

To find the number of local maximum and minimum points for \( f(x) = \int_0^{x^2} \frac{t^2 - 8t + 15}{e^t} dt \), we analyze its derivative. Using the Leibniz rule, the derivative is:
\[ f'(x) = \frac{d}{dx}\left(x^2\right) \cdot \frac{x^2 - 8x + 15}{e^{x^2}} = 2x \cdot \frac{x^2 - 8x + 15}{e^{x^2}} \]
Setting \( f'(x) = 0 \) yields:
\[ 2x(x^2 - 8x + 15) = 0 \]The solutions are:

• \( 2x = 0 \Rightarrow x = 0 \)
• \( x^2 - 8x + 15 = 0 \Rightarrow (x-3)(x-5) = 0 \Rightarrow x = 3, 5 \)

A sign test on \( x(x-3)(x-5) \) reveals the intervals of increase and decrease:

• \( x < 0 \): \( f'(x) > 0 \) (increasing)
• \( 0 < x < 3 \): \( f'(x) < 0 \) (decreasing)
• \( 3 < x < 5 \): \( f'(x) > 0 \) (increasing)
• \( x > 5 \): \( f'(x) < 0 \) (decreasing)

Local extrema occur at sign changes of the derivative. Thus:

• Local maxima are at \( x = 0, 5 \)
• Local minimum is at \( x = 3 \)

Considering points beyond 5, the counts of local maxima and minima are 2 each.