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 Calculus Calculus Math Forum

 August 17th, 2016, 03:37 PM #1 Newbie   Joined: May 2013 Posts: 19 Thanks: 0 Critical Points Hello I'm calculating critical points, so it's f'(x)=0 f'(x)= (4/3)x- tanx then (4/3)x-tanx=0 Wolfram alpha states that one of the answers is +-(0.844) I tried rewriting it as: (4/3)xcosx-sinx=0 ; by rewriting tanx as sinx/cosx but I couldn't think of a numerical solution. Is there any way to solve this using algebra? or any trig trick that I may be missing. Thanks  August 17th, 2016, 04:17 PM   #2
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$f'(x) = \dfrac{4}{3}x - \tan{x} = 0$

$f'(x)$ is an odd function, so zeros will be symmetrical to the y-axis.

$x = 0$ is a solution ... any others will require a numerical method
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