Inequality Derivative

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racuna

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Sep 28, 2005
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My problem is: f(x)=8x-cosx,0<=x<=2pi
It asks on what interval(s) is f concave downward?

:p I know that in order to find the concavity of the function I need to find the second derivative, but I don't really understand how to start the differentiation. Please help!
 
Derive it as normal. Just remember that f is only defined on [0,2pi], so your answer will be an inveral on [0,2pi]
 
How to start the differentiation.

f(x) = 8x-cosx ,0<=x<=2pi
f'(x)= 8+sinx
f''(x)= cosx

ps: defferentiation of nx=x and defferentiation of n=0
 
f(x) = 8x-cosx ,0<=x<=2pi
f'(x)= 8+sinx
f''(x)= - cosx

close. (sin(x))' = cos(x) not -cos(x) though.
so f''(x) = cos(x). From there you can figure out concavity by methods outlined in your textbook.
 
yeah.. u right. Mistyped :lol: the derivative of sinx= cox, but the derivative of cosx= - sinx
 
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