Trace Mcintyre

2022-05-03

I have the following question:

Given a polynom $p(x)$ I need to prove the the equation $\mathrm{tan}x=p(x)$ has at least one solution.

I defined the difference as a function $f(x)=\mathrm{tan}x-p(x)$ and I know that $f(x)$ is continues in any interval where $\mathrm{tan}x$ is defined, so I am trying to use the Intermediate value theorem, but I don't know how.

Given a polynom $p(x)$ I need to prove the the equation $\mathrm{tan}x=p(x)$ has at least one solution.

I defined the difference as a function $f(x)=\mathrm{tan}x-p(x)$ and I know that $f(x)$ is continues in any interval where $\mathrm{tan}x$ is defined, so I am trying to use the Intermediate value theorem, but I don't know how.

RormFrure6h1

Beginner2022-05-04Added 13 answers

Consider

$g(x)=\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}p(x)-\mathrm{sin}x$

We have $g(-\pi /2)=1$ and $g(\pi /2)=-1.$ Hence $g({x}_{0})=0$ for a point ${x}_{0},$ $-\pi /2<{x}_{0}<\pi /2.$ Thus

$p({x}_{0})=\frac{\mathrm{sin}{x}_{0}}{\mathrm{cos}{x}_{0}}=\mathrm{tan}{x}_{0}$

$g(x)=\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}p(x)-\mathrm{sin}x$

We have $g(-\pi /2)=1$ and $g(\pi /2)=-1.$ Hence $g({x}_{0})=0$ for a point ${x}_{0},$ $-\pi /2<{x}_{0}<\pi /2.$ Thus

$p({x}_{0})=\frac{\mathrm{sin}{x}_{0}}{\mathrm{cos}{x}_{0}}=\mathrm{tan}{x}_{0}$

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