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: 2,977 Thanks: 1571 
$f'(x) = \dfrac{4}{3}x  \tan{x} = 0$ $f'(x)$ is an odd function, so zeros will be symmetrical to the yaxis. $x = 0$ is a solution ... any others will require a numerical method 

Tags 
critical, points 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Critical points, oh please help on this  irvm  Calculus  6  January 15th, 2016 05:54 AM 
Critical Points  mathe  Calculus  2  November 11th, 2015 12:32 PM 
Critical Points  mathkid  Calculus  1  November 11th, 2012 06:34 PM 
critical points  summerset353  Calculus  1  March 5th, 2010 01:50 AM 
critical points of (24x^2)/(1+x^2)^2  stainsoftime  Calculus  3  November 24th, 2008 04:24 AM 