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prove that there is some number x such that $ x^{179}+\frac{163}{1+x^2+\sin^2x}=119$

I really want to know the general process to prove something like that...

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2 Answers

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First notice $1+x^2+\sin^2(x)\geq 1$ so the denominator is never $0$ and the function $f(x)=x^{179}+\frac{163}{1+x^2+\sin^2(x)}-119$ is defined and continuous everywhere.

$f(2)>2^{179}-119>0$, $f(-2)<-2^{179}+163-119<0$ so by intermediate value theorem there is a $x_0\in (-2,2)$ satisfies $f(x_0)=0$, which is the solution of the equation given.

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The dominating term is $x^{179}$, so as $x\to\pm\infty$, $f(x)\to\pm\infty$. In particular, the intermediate Value Theorem guarantees the existence of at least one $x$ such that $f(x)=c$ for any given $c\in\Bbb R$.

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