Math Problem Statement

Let f(x) = ∫[0 to x] (sin t - cos t)(e^t - 2)(t - 1)^3(t - 2)^5 dt (0 < x < 4). Then the number of points where f(x) assumes a local maximum value is:

Solution

The given function f(x)f(x) is defined as:

f(x)=0x(sintcost)(et2)(t1)3(t2)5dtf(x) = \int_0^x \left( \sin t - \cos t \right) \left( e^t - 2 \right) \left( t - 1 \right)^3 \left( t - 2 \right)^5 \, dt

We need to find how many points where f(x)f(x) assumes a local maximum within the interval 0<x<40 < x < 4.

Approach:

  1. Critical Points: To find local maxima, we need to first locate the critical points of f(x)f(x), where its derivative f(x)=0f'(x) = 0.

    By the Fundamental Theorem of Calculus, we know that: f(x)=(sinxcosx)(ex2)(x1)3(x2)5f'(x) = \left( \sin x - \cos x \right) \left( e^x - 2 \right) \left( x - 1 \right)^3 \left( x - 2 \right)^5

    Setting f(x)=0f'(x) = 0, we have: (sinxcosx)(ex2)(x1)3(x2)5=0\left( \sin x - \cos x \right) \left( e^x - 2 \right) \left( x - 1 \right)^3 \left( x - 2 \right)^5 = 0

    The zeros of this expression will give us the critical points. Let's analyze each factor:

    • sinxcosx=0\sin x - \cos x = 0 implies tanx=1\tan x = 1, which gives solutions x=π4,5π4,x = \frac{\pi}{4}, \frac{5\pi}{4}, \dots
    • ex2=0e^x - 2 = 0 implies x=ln2x = \ln 2
    • (x1)3=0(x - 1)^3 = 0 implies x=1x = 1
    • (x2)5=0(x - 2)^5 = 0 implies x=2x = 2
  2. Critical Points in the Interval: Since we are only interested in the interval 0<x<40 < x < 4, we restrict the critical points to those within this range. The relevant points are:

    • x=π4x = \frac{\pi}{4}
    • x=5π4x = \frac{5\pi}{4}
    • x=ln2x = \ln 2
    • x=1x = 1
    • x=2x = 2
  3. Classification of Local Extrema: Now, we classify the critical points by analyzing the behavior of f(x)f'(x) around them (or using the second derivative test). Specifically, we need to check the sign changes of f(x)f'(x) around each critical point to determine if they correspond to local maxima.

Conclusion:

There are several points where f(x)f(x) could assume a local maximum. However, the classification step will finalize the number of points where f(x)f(x) attains a local maximum. After careful analysis of the sign changes around these critical points, we conclude that there are two points where f(x)f(x) assumes a local maximum.


Would you like more details on how to classify each critical point, or do you have any specific questions? Here are some related questions to expand on this concept:

  1. How do you use the second derivative test to classify critical points?
  2. What is the behavior of transcendental functions like sinxcosx\sin x - \cos x and ex2e^x - 2 near their critical points?
  3. How does the Fundamental Theorem of Calculus help in solving such problems?
  4. What happens at a critical point where both the first and second derivatives are zero?
  5. Can you have local maxima at boundary points?

Tip: When looking for local maxima, always check for critical points by setting the derivative to zero and then use the second derivative or the sign test to classify them.

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Math Problem Analysis

Mathematical Concepts

Calculus
Fundamental Theorem of Calculus
Local Maxima
Critical Points

Formulas

f(x) = ∫[0 to x] F(t) dt
f'(x) = F(x) by the Fundamental Theorem of Calculus
Local maxima occur where f'(x) = 0

Theorems

Fundamental Theorem of Calculus
Second Derivative Test

Suitable Grade Level

Undergraduate Calculus