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I want to prove that $2^x=2-x^2$ has at least two real solutions. I am trying to use Bolzano theorem to prove this, that is if $f(x)=2^x-2+x^2$=$0$ then $f(x)$ is continuous on R will be negative somewhere and positive somewhere and so satisfy Bolzano theorem and there will be some real value that $f(c)=0$, but I don't know how to show or prove this equation has at least two real solutions.

I appreciate anyone who show that for me.

Thank you.

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Try $x=0$, $x=1$, and $x=-2$ as inputs for $f$.

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  • $\begingroup$ Thanks Jonas. Well that is what I did after I Plotted the function in Maple and I noticed where is the function negative and where is positive, but I want to be more precisely and to get more precisely proof if it is possible ( I mean how to prove using the principals of analysis maybe?) $\endgroup$ – LoveMath May 1 '13 at 8:38
  • $\begingroup$ @user50382: Use the intermediate value theorem twice. $\endgroup$ – Jonas Meyer May 1 '13 at 11:39

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