Concavity of y = x(cosx) at x = pi/3: Second Derivative Test

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To determine the concavity of the function y = x(cosx) at x = pi/3 using the second derivative test, the first derivative is calculated as y' = -xsinx + cosx, and the second derivative is y'' = -xcosx - 2sinx. The next step involves solving the equation -xcosx = 2sinx, which is necessary to find the concavity. By substituting x = pi/3 into the second derivative, the concavity can be assessed. The discussion highlights the importance of correctly applying the second derivative test for concavity analysis.
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I am doing this problem and I am getting stuck at solving the equation

problem: Use the second derivative test to determine the concavity of the following function. y = x(cosx) at x = pi/3

solution: y' = -xsinx + cosx
y'' = -xcosx - 2sinx = 0

and then i did
-xcosx = 2sinx ( i don't know if this is correct)
and then I am stuck... i don't know how to proceed.

please help
 
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The function is upward concave is f''(x) > 0, and downward concave if f''(x) < 0. So just plug x=pi/3 into your function for y''.
 
ohhhh! duhhh! I shouldve known... thanks
 
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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