Finding relative extrema of an integral function f(x) = ∫(t² − 4)/(1 + cos²t)dt

[science.girl]

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?)

 

[ideasrule]

Your work is right; deriving an integral yields the original equation inside the integral.

 

[science.girl]

Your work is right; deriving an integral yields the original equation inside the integral.

Thank you for the clarification. =)

 

[jbunniii]

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.

 

[science.girl]

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!