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funky question (1 Viewer)

kwabon

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just wondering how would you prove this ALGEBRAICALLY, i could do it graphically not algebraically. :-S, prove; sin x - x + (1/6)x^3 >= 0 for x > 0
 

darusan92

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just wondering how would you prove this ALGEBRAICALLY, i could do it graphically not algebraically. :-S, prove; sin x - x + (1/6)x^3 >= 0 for x > 0
umm.... try differentiating it n prove that it has a positive gradient for x>0 i think they both are the same algebracally and graphically (proving positive gradient)??? just check it in case.....
 

The Nomad

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Check the inequalities exercise in Cambridge, harder 3U chapter. The last few questions are similar to this if I remember correctly. If not, I'm sure one of the examples is similar to this question.
 

Trebla

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Trebla

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Edited.
 
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lolokay

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just wondering how would you prove this ALGEBRAICALLY, i could do it graphically not algebraically. :-S, prove; sin x - x + (1/6)x^3 >= 0 for x > 0
differentiate the function, f(x) = sin x - x + (1/6)x^3
f'(x) = cosx - 1 + x^2/2
this is equal to zero when cosx = 0, so this is the minimum or maximum value f(x) in this range
and f''(x) = -sinx + x > 0 for x >0
so this is the minimum value
.'. f(x) > f(0) for x > 0
sin x - x + (1/6)x^3 > 0
 
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