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Could someone explain how to solve this problem? Im totally stuck!

 Sep 12, 2019
edited by Soapen  Sep 13, 2019
 #1
avatar+103858 
+2

I think that the  inequality  is incorrect

 

See the graph here :  https://www.desmos.com/calculator/alijngoclf

 

Notice that x^2   is  ≥ sin^2x    everywhere on   (-pi/2, pi/2)

 

But  if     x^2  ≤ f(x)   on this interval...then f(x) is ≥ x^2  everywhere on this interval...then f(x)  cannot also be  ≤ sin^2x  everywhere on the interval

 

I believe that the inequality we need is  this

 

sin^2 x  ≤ f(x)   ≤ x^2

 

Note that if this true......then

 

f(x)                 f(x)      =    0          by the Squeeze Theorem

lim x → 0   

 

 

 

cool cool cool   

 Sep 12, 2019
 #2
avatar+12 
0

Thank u!!!

Soapen  Sep 13, 2019

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