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

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SUMMARY

The discussion focuses on determining the concavity of the function y = x(cosx) at x = π/3 using the second derivative test. The first derivative is calculated as y' = -xsinx + cosx, while the second derivative is y'' = -xcosx - 2sinx. To find concavity, the value of y'' at x = π/3 must be evaluated. The function is upward concave if f''(x) > 0 and downward concave if f''(x) < 0, leading to the conclusion that substituting x = π/3 into y'' will clarify the concavity.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives.
  • Familiarity with the second derivative test for concavity.
  • Knowledge of trigonometric functions, particularly sine and cosine.
  • Ability to evaluate functions at specific points, such as x = π/3.
NEXT STEPS
  • Calculate the value of y'' at x = π/3 to determine concavity.
  • Review the second derivative test and its applications in calculus.
  • Explore the behavior of trigonometric functions in calculus.
  • Practice additional problems involving concavity and inflection points.
USEFUL FOR

Students studying calculus, particularly those learning about derivatives and concavity, as well as educators seeking to clarify the second derivative test for their students.

donjt81
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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
 

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