
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 14th, 2017, 08:25 PM  #1 
Newbie Joined: Nov 2016 From: USA Posts: 26 Thanks: 1  Please clarify the final step of this proof (inflection point)
Here is the solution to proving the inflection point of y=xsinx lies on the curve y^2(x^2+4)=4x^2 y'' 2cosx  xsinx = 0 xsinx=2cosx (hereafter all are x's are x sub zero) inflection point of the curve (x, 2cos x) Prove (x,2cosx) lies on the curve y^2(x^2+4)=4x^2 (2cosx)^2(x^2+4)=4x^2 4cos^2x[(4 cos^2x/sin^2x) +4] = 4[(2cosx)/sinx]^2 16 (cos^2x/sin^2x) = 16 (cos^2x/sin^2x) How does the first bolded equation equal the second bolded equation? 
April 15th, 2017, 05:05 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 680 
4cos^2(x)/sin^2(x)+ 4= 4((cos^2(x)/sin^2(x))+ 1)= 4(cos^2(x)sin^2(x)+ sin^2(x)/sin^2(x))= 4((cos^2(x)+ sin^2(x))/sin^2(x))= 4/sin^2(x). Multiplying that by 4 cos^2(x) gives 16 cos^2(x)/sin^2(x). 
April 15th, 2017, 06:57 AM  #3  
Member Joined: Jan 2016 From: Athens, OH Posts: 46 Thanks: 26 
First comment: I don't like the way you started the proof. Quote:
The following graphs show the result and also the fact that there are infinitely many inflection points. But also notice not every intersection of the two graphs is an inflection point.  

Tags 
clarify, final, inflection, point, proof, step 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
how did they come to the final step of this equation?  vlekje5  PreCalculus  3  April 12th, 2017 11:32 AM 
A point of inflection is a turning point or not  helloprajna  Calculus  5  February 6th, 2015 02:57 AM 
Inflection point  paper006  Calculus  1  June 2nd, 2014 11:31 AM 
Point of Inflection  JenniferC  Calculus  2  January 23rd, 2014 04:12 PM 
Inflection point  ungeheuer  Calculus  2  August 21st, 2013 02:29 PM 