Yahoo Answers: Answers and Comments for How do you solve an equation in the form (x^2+x+1)*e^x<1/100? [Mathematics]
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From Anonymous
enUS
Fri, 04 Feb 2011 21:40:42 +0000
3
Yahoo Answers: Answers and Comments for How do you solve an equation in the form (x^2+x+1)*e^x<1/100? [Mathematics]
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From Anonymous: As x is negative and increasing in absolute va...
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https://answers.yahoo.com/question/index?qid=20110204214042AAOWUPl
Sat, 05 Feb 2011 07:28:21 +0000
As x is negative and increasing in absolute value, e^x gets tiny.
Foe example e^(9) = 0.0001341
Just for fun, let's check x = 9 out.
(81  9 + 1) = 73
73 * e^(9) = 0.00901 < (1 / 100)
Since the derivative of (x^2+x+1)*e^x is (2x+1)*e^x + (x^2 + x + 1)*e^x =
(x^2 + 3x + 2) * (e^x) it is clear that this derivative is always positive and thus that the function is always increasing.
As 9 almost provided equality let's try x = 8.5
That gets me 0.01328 > 0.01
So the answer lies between 8.5 and 9
Hence since the inequality is true for x = 9, it is true for x < 9
It looks closer to equality so I'll try x =  8.862. That gets me 0.01001 That's really close
Last try: x =  8.8625. That gets 0.000141416
That is very, very close too 0.01 and less than 0.01
So while this was done by approximation (and the fact that the derivative is positive) I'm going to go with a final answer of:
x =  8.8625
.