Using mean value theorem , prove that $\displaystyle tan x > x$ for all $\displaystyle x \boldsymbol{\epsilon} ( 0, \frac{\pi}{2} )$
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For all x in (0,pi/2) there is a c in (0,x) such that [tan(x)-tan(0)]/[x-0] = sec^2(c) > 1 for c in (0,pi/2) tan(x)/x > 1 tan(x) > x
Originally Posted by Calculus26 For all x in (0,pi/2) there is a c in (0,x) such that [tan(x)-tan(0)]/[x-0] = sec^2(c) > 1 for c in (0,pi/2) tan(x)/x > 1 tan(x) > x I didnt get this part could u please explain [tan(x)-tan(0)]/[x-0] = sec^2(c)
Originally Posted by zorro I didnt get this part could u please explain [tan(x)-tan(0)]/[x-0] = sec^2(c) that is just applying the mean value theorem. [f(b)-f(a)] / (b-a) = f'(c), in this case a=0, b=x, and f(x) = tan(x) so f'(x) = sec^2(x).
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