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 identify the number of local maxima and minima for the function \( f(x) \), we first compute its first and second derivatives. Differentiating \( f(x) \) with respect to \( x \) using the Leibniz rule for the derivative of an integral yields: \[ f'(x) = \frac{d}{dx} \left( \int_0^{x^2} \frac{t^2 - 8t + 15}{e^t} dt \right) = \frac{d}{dx} \left( \int_0^{x^2} g(t) dt \right), \] where \( g(t) = \frac{t^2 - 8t + 15}{e^t} \). Applying the Leibniz rule, we obtain: \[ f'(x) = 2x \cdot \frac{(x^2)^2 - 8(x^2) + 15}{e^{x^2}}. \] Critical points are found by setting \( f'(x) = 0 \): \[ 2x \cdot \frac{x^4 - 8x^2 + 15}{e^{x^2}} = 0. \] This equation is satisfied if \( x = 0 \) or \( x^4 - 8x^2 + 15 = 0 \). Solving the quadratic equation in \( x^2 \), \( x^4 - 8x^2 + 15 = 0 \), gives: \[ x^2 = \frac{8 \pm \sqrt{64 - 60}}{2} = 4 \pm \sqrt{1}, \] resulting in \( x^2 = 5 \) and \( x^2 = 3 \). Therefore, the solutions are \( x = \pm \sqrt{5} \) and \( x = \pm \sqrt{3} \).

We then analyze the nature of these critical points using the second derivative \( f''(x) \). By calculating \( f''(x) \) and evaluating its sign at the critical points \( x = 0, \pm \sqrt{3}, \pm \sqrt{5} \), we determine that there are 2 local maxima and 2 local minima. 

Consequently, the function has 2 local maxima and 2 local minima.