Real roots of 2^x=1+x^2

[kinu]

the number of real roots of the equation \(2^x = 1+x^2\)

[josesousa]

We already solved similar problems in here. Didn't it help?

[João P. Ferreira]

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

[kinu]

Thanks Sir