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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.

LoveMath
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2 Answers2

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

Jonas Meyer
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  • 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?) – LoveMath May 01 '13 at 08:38
  • @user50382: Use the intermediate value theorem twice. – Jonas Meyer May 01 '13 at 11:39
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Let $f$ and $g$ by the functions such that $f(x)=2^x+x^2-2$ for $x \in [-2,0]$ and $g(x)=2^x+x^2-2$ for $x \in [0,1]$.

Nos, apply Bolzano’s Theorem on both functions. This gives you (at least) to solutions to the equation.

James Garrett
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