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x^2+x>1/e^4

 Dec 26, 2015

Best Answer 

 #3
avatar+104793 
+10

x^2+x >1/e^4        make this an equality and complete the square

 

x^2 + x + 1/4 = 1/e^4 + 1/4       factor the left side, simplify the right

 

(x + 1/2)^2  = [ 4 + e^4] / [4e^4)   take the square root of both sides

 

x + 1/2  =  +/- sqrt [ 4 + e^4] / [2e^2]

 

x = +/- sqrt [ 4 + e^4] / [ 2e^2] - 1/2  ≈  [ -1.0179919293664084, 0.0179919293664084 ]

 

The answer  will come from one or more of these intervals

 

(-infinity, - sqrt [ 4 + e^4] / [ 2e^2] - 1/2) , (- sqrt [ 4 + e^4] / [ 2e^2] - 1/2,  sqrt [ 4 + e^4] / [ 2e^2] - 1/2), ( sqrt [ 4 + e^4] / [ 2e^2] - 1/2, infinity)

 

Choosing 0 as a test point for the middle interval will make the inequality false

 

So.....the two outside intervals solve the inequality

 

Here's the graph that proves this........https://www.desmos.com/calculator/iluvf1vudp

 

 

 

cool cool cool

 Dec 26, 2015
 #1
avatar
0

x^2+x>1/e^4

 

=x>sqrt(4+e^4)/(2e^2)-1/2

 Dec 26, 2015
 #2
avatar
+5

 in last link you will find your answer by stepwise 

https://www.symbolab.com 

 Dec 26, 2015
 #3
avatar+104793 
+10
Best Answer

x^2+x >1/e^4        make this an equality and complete the square

 

x^2 + x + 1/4 = 1/e^4 + 1/4       factor the left side, simplify the right

 

(x + 1/2)^2  = [ 4 + e^4] / [4e^4)   take the square root of both sides

 

x + 1/2  =  +/- sqrt [ 4 + e^4] / [2e^2]

 

x = +/- sqrt [ 4 + e^4] / [ 2e^2] - 1/2  ≈  [ -1.0179919293664084, 0.0179919293664084 ]

 

The answer  will come from one or more of these intervals

 

(-infinity, - sqrt [ 4 + e^4] / [ 2e^2] - 1/2) , (- sqrt [ 4 + e^4] / [ 2e^2] - 1/2,  sqrt [ 4 + e^4] / [ 2e^2] - 1/2), ( sqrt [ 4 + e^4] / [ 2e^2] - 1/2, infinity)

 

Choosing 0 as a test point for the middle interval will make the inequality false

 

So.....the two outside intervals solve the inequality

 

Here's the graph that proves this........https://www.desmos.com/calculator/iluvf1vudp

 

 

 

cool cool cool

CPhill Dec 26, 2015

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