For 0<x<180 verify the monotony of the function xcosx+sin(-x).
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We have the function f(x) = x*cos x + sin (-x)
f(x) = x*cos x - sin x
The first derivative of f(x) is f'(x)
=> f'(x) = cos x - x*(sin x) - cos x
=> f'(x) = -x*sin x
When 0< x< 180 we have sin x as positive, therefore the first derivative -x*sin x is negative, so the function is decreasing.
Therefore the function is decreasing for the given range of x.
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The monotony of a function shows the behavior of the function: increasing or decreasing function.
A function is strictly increasing if it's first derivative is positive and it is decreasing if it's first derivative is negative.
We'll re-write the function, based on the fact that the sine function is odd:
f(x) = xcos x - sin x
We'll calculate the first derivative:
f'(x)= (xcos x – sin x)'
f'(x) = ( xcos x)'-( sin x)'
We notice that the first term is a product, so we'll apply the product rule:
f'(x) = x'*(cos x)+x*(cos x)' – cos x
f'(x)=1*cos x-x*sin x –cos x
We'll eliminate like terms:
f'(x)= -x*sin x
Since the sine function is positive over the range [0 ; 180], the values of x are positive and the product is negative, the first derivative is negative.
The function y = f(x) = xcosx+sin(-x) is decreasing over the range [0, pi].
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