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Let $p(x)$ be a given polynomial. Show that there is a constant $K$ such that the function $g$ defined by

$g(x)=\ln(p(x)+K)$

is uniformly continuous on the interval $[-10,10]$

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  • $\begingroup$ If you make it continuous, by Cantor's theorem it will be uniformly continuous. To make it continuous just make sure that $p(x)+K>0$ on $[-10,10]$. Since $p$ is continuous it attains a minimum $m$ on $[-10,10]$. You could take $K=\epsilon+\min(0,m)$ for some $\epsilon>0$. $\endgroup$ – conditionalMethod Nov 13 at 20:26
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Every continuous function on a compact set is uniformly continuous. However, not all values of $K$ will give you a function that is defined on the whole interval, which is the obstruction here. The uniform condition is a red herring.

If $p(x) $ is always positive you can set $K=0$. Otherwise, let $M$ be the minimum value of $p$ and set $K=|M|+1$. Then $p(x)+K$ is always positive, so we have a continuous function on $[-10,10]$.

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$p$ as polynomial is continuous at the compact $ [-10,10] $, thus $ p $ is bounded. This means that there exists $ K>0$ such that $$\forall x\in[-10,10] \;\; -K<p(x)<K$$ hence the function $$x\mapsto \ln(p(x)+K)$$ is well defined and continuous at $ [-10,10]$ and by Heine's theorem, it is uniformly continuous at $ [-10,10]$.

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Observe that $p$ is continuous, and hence there exist $m\le M$ such that $$ m\le p(x)\le M, \quad \text{for all $x\in [-10,10]$.} $$ Hence, clearly $$ p(x)-m+1\ge 1, \quad\text{for all $x\in [-10,10]$.} $$ Thus $$ f(x)=\ln\big(p(x)-m+1\big) $$ is well defined and continuous in $[-10,10]$, and hence uniformly continuous, since $[-10,10]$ is compact, i.e., closed and bounded subset of $\mathbb R$.

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