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let $f: \Bbb R \to \Bbb R $ be defined by $f(x) = x^5 + 2x^3 + 3x^2 + 6$

Prove that function is continuous?

I am just stuck on the first step how to manipulate the function algebraically and then I can apply the $\epsilon - \delta$ definition. Any hints on how to that?? Thanks in advance!!

Shervan
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2 Answers2

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OK, here is an explicit proof working from the definition and only the definition.

We show $f$ continuous on $\Bbb R$. (We have to show it equivalently at any point.)

Let $a$ be a point in $\Bbb R$, we show the continuity in $a$. We use as a notation $b$ for $b:=f(a)$.

Let us give us an $\epsilon$-ball around $b$, $\epsilon>0$ fixed now. We may and do assume $\epsilon < 1$. (Else we replace it with $1$.)

(For it we have to find a $\delta>0$, so that for all $x$ at distance $<\delta$ from $a$ the mapped point $f(x)$ is at distance $\epsilon$ from $b$.) Note that we have the brute estimation: $$ \begin{aligned} |f(x)-f(a)| &= |\ (x^5+2x^3+3x^2+6)-(a^5+2a^3+3a^2+6)\ | \\ &\le |x^5-a^5|+2|x^3-a^3|+3|x^2-a^2| \\ &=|x-a|\;|x^4+x^3a+x^2a^2+xa^3+a^4| \\ &\qquad+2|x-a|\;|x^2+xa+a^2| \\ &\qquad\qquad+3|x-a|\;|x+a| \\ &\le |x-a|\;\Big( \ |x|^4+|x|^3|a|+|x|^2|a|^2+|x||a|^3+|a|^4\ \Big) \\ &\qquad+2|x-a|\Big(\ |x|^2+|x||a|+|a|^2\ \Big) \\ &\qquad\qquad+3|x-a|\Big(\ |x|+|a|\ \Big) \\ &\le |x-a|\Big(\ (|x|+|a|)^4+2(|x|+|a|)^2+3(|x|+|a|)\ \Big) \\ &\le |x-a|\underbrace{\Big(\ (|a|+1+|a|)^4+2(|a|+1+|a|)^2+3(|a|+1+|a|)\ \Big)} _{=:M}\ . \end{aligned} $$ We use now the constant $M>0$ from above to define $\delta:=\epsilon/M$.

Let now $x$ be so that $|x-a|<\delta$.

Then we have $|f(x)-f(a)|\le M\; |x-a|<M\delta=\epsilon$.

$\square$

dan_fulea
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  • One can also factor the polynomial, to get a shorter estimation, but it is better to write things so that they apply in general. – dan_fulea Nov 15 '19 at 19:19
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If you already know the following, proving your polynomial(or any other one) is continuous is very easy:

$f(x) = c$ is continuous for a constant $c$,

$f(x) = x$ is continuous,

If $f(x)$ and $g(x)$ are continuous, then so are $f(x)g(x)$ and $f(x) + g(x)$