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| Real roots of 2^x=1+x^2 https://forumdematematica.org/viewtopic.php?f=72&t=167 |
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| Autor: | kinu [ 27 jan 2012, 16:51 ] |
| Título da Pergunta: | Real roots of 2^x=1+x^2 |
the number of real roots of the equation \(2^x = 1+x^2\) |
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| Autor: | josesousa [ 27 jan 2012, 20:13 ] |
| Título da Pergunta: | Re: Real roots |
We already solved similar problems in here. Didn't it help? |
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| Autor: | João P. Ferreira [ 27 jan 2012, 23:29 ] |
| Título da Pergunta: | Re: Real roots |
Kinu Check this We can easily see that \(x=0\) and \(x=1\)are solutions \(2^{0}=1+0^{2}\) and \(2^{1}=1+1^{2}\) When \(x<0\) we can see that there is no roots cause\(2^x<1+x^2, \ x<0\) The rightest root we know by now is \(x=1\) We can find the derivative at \(x=1\) for both funtions \(f(x)=2^x\) and \(g(x)=1+x^2\) \(f'(1)=\frac{2}{ln(2)} \ \ g'(1)={1}+{2}={3}\) We know that \(\frac{2}{ln(2)}<3\) and we also know that \(\lim_{x \to +\infty}\frac{2^x}{1+x^2}=+\infty\) which means for every \(x>0\) there is \(y>x\ : \ 2^y>x^2+1\) which means that there is three roots As Prof. José Sousa said there is a similar topic here Take care kinu |
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| Autor: | kinu [ 28 jan 2012, 05:09 ] |
| Título da Pergunta: | Re: Real roots of 2^x=1+x^2 |
Thanks Sir |
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