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If n is a positive integer, how many times does the function f(x)=x^2 + 5cosx change concavity in the interval?
0 less than or equal to x less than or equal to 2(pi)*n
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- Anonymous1 decade agoFavorite Answer
f '(x) = 2x - 5*sin(x)
Set this equal to 0
x = ((5*sin(x)) / 2)
Take the second derivative.
f"(x) = -5*cos(x)
This will be a point of inflection for x = π/2, 3π/2, 5π/2...
So there will be a point of inflection every π increase. For 2π that will be 2, and for 2π*n there will be 2n.
Answer: 2n
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