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# How do you solve an equation in the form (x^2+x+1)*e^x<1/100?

I'm not sure how to solve this when you include the e^x. Any suggestions?

### 1 Answer

- Anonymous9 years agoFavorite Answer
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

.

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