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)xtanx=0 Wolfram alpha states that one of the answers is +(0.844) I tried rewriting it as: (4/3)xcosxsinx=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 
Math Team Joined: Jul 2011 From: Texas Posts: 3,031 Thanks: 1620 
$f'(x) = \dfrac{4}{3}x  \tan{x} = 0$ $f'(x)$ is an odd function, so zeros will be symmetrical to the yaxis. $x = 0$ is a solution ... any others will require a numerical method 

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