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April 14th, 2017, 08:25 PM  #1 
Member Joined: Nov 2016 From: USA Posts: 33 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: 2,978 Thanks: 807 
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: 83 Thanks: 41 
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.  

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