Find Relative Extrema of f(x) | Maxima/Minima Question Homework

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Homework Help Overview

The problem involves finding the relative extrema of the function defined by an integral, specifically f(x) = ∫(t² - 4)/(1 + cos²(t)) dt from 0 to x. Participants are discussing the behavior of the derivative of this function to identify maxima and minima.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the critical points by setting the derivative f'(x) to zero, leading to the values x = ±2. They question whether both points are minima and if further differentiation is necessary.
  • Some participants confirm the correctness of deriving the integral but raise concerns about the interpretation of the behavior of f'(x) at the critical points.
  • Others suggest reconsidering the implications of f'(x) being an even function, which affects the analysis of the slopes at the critical points.

Discussion Status

Contextual Notes

Participants are navigating the implications of the function's evenness and its effect on the analysis of extrema, which may influence their conclusions about the nature of the critical points.

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Homework Statement


Find relative extrema of f(x).

f(x) = \int^{x}_{0} (t^{2} -4)/(1 + cos(t)^{2})

Homework Equations


N/A

The Attempt at a Solution


Is this correct?

f '(x) = [(x² - 4)/(1 + cos²x)]

Now set f '(x) = 0,
[(x² - 4)/(1 + cos²x)] = 0
x² - 4 = 0
x = ± 2

f'(x) is changing from negative to positive for both +2 and -2, so are both of them minima?
(And would you have to take the derivative of the so-called f'(x) to get the actual derivative from which to calculate the max/min?)
 
Last edited:
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Your work is right; deriving an integral yields the original equation inside the integral.
 
ideasrule said:
Your work is right; deriving an integral yields the original equation inside the integral.

Thank you for the clarification. =)
 
science.girl said:
f'(x) is changing from negative to positive for both +2 and -2

No, that's not true. f'(x) is an even function of x, so whatever its slope is at -2 must be the negative of its slope at 2.
 
jbunniii said:
No, that's not true. f'(x) is an even function of x, so whatever its slope is at -2 must be the negative of its slope at 2.

Ah; makes sense. Thank you for pointing this out!
 

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