I am not sure how to prove rigorously (using calculus) that $x^2=2^x$ has exactly $3$ real solutions.
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You can study the continuous function defined on $\mathbb{R}$ by $f(x)=x^2-2^x=x^2-e^{x\ln(2)}$ (using derivative and second derivative). – Augustin Aug 22 '15 at 22:26
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@Augustin you mean $e^{x\ln(2)}$... – YoTengoUnLCD Aug 22 '15 at 22:28
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Yes thanks, I edit. – Augustin Aug 22 '15 at 22:29
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Let $f(x)=2^x-x^2$. Then, $f'''(x)$ is always positive, so $f$ has at most three roots by Rolle's theorem. To prove that there are at least three roots, note that $f(-1)=-0.5$, $f(0)=1$, $f(3)=-1$, $f(5)=7$.
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