Finding the absolute maximum and absolute minimum

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To find the absolute maximum and minimum of the function f(t) = t + cot(t/2) on the interval [π/4, 7π/4], the derivative has been calculated as 1 - (1/2)csc²(t/2). The next step involves identifying critical points by solving the equation derived from the derivative. Progress has been made in determining that sin(t/2) equals ±1/√2, which corresponds to specific angles. The discussion emphasizes the importance of graphing sin(x) to locate these angles accurately for further analysis.
meeklobraca
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Homework Statement



Find the absolute max and min of

f(t)=t + cot(t/2) on [pi/4, 7pi/4]

Homework Equations





The Attempt at a Solution



I have attempted to find the derivative which I believe is

1 - csc^2 (t/2) * (t/2) which I can simplify down to cot^2 (t/2) * (t/2)

Even if that is correct, which I am doubtful of, where do I go from here?

Thank You!
 
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Why do you think the derivative of t-cot(t/2) is 1-(csc^(t/2))*(t/2)? Better fix that before you try to proceed.
 
Okay I've found the derivative of t + cot (t/2) to be

1 - csc^2(t/2) * 1/2
 
meeklobraca said:
Okay I've found the derivative of t + cot (t/2) to be

1 - csc^2(t/2) * 1/2

That's much better. So do you have any critical points on [pi/4, 7pi/4]?
 
Here in lies the problem Dick. Where do you go from here?

im at sqrt 2 = csc (t/2) from which I've turned into sqrt 2 = 1/sin (t/2)
 
That's pretty good progress. Actually possibilities are at sin(t/2)=+/-1/sqrt(2). Can you draw a graph of sin(x) and tell me where sin(x)=+/-1/sqrt(2)? Then put t/2=x. I happen to know sin(pi/4)=1/sqrt(2).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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