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$f(x) = \dfrac{x^7 + x^5 + x^3 - x^2 - 1}{6}$

I'm not getting any idea how to find the range of this function. I know that to find the range of a function, we find the range of inverse function. But in the case of this function, that's quite not possible maybe. By seeing the graph one can conclude that range is $(-\infty, \infty)$ But how can we find range without looking at the graph?

I try the following method.

$f(x) = \dfrac{x^5(x^2 + 1) + x^3 -1(x^2+1)}{6}$

$f(x) = \dfrac{(x^5-1)(x^2 + 1) + x^3}{6}$

Can we conclude any result from here?

Is there any other more fundamental method to solve such types of questions?

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    $\begingroup$ It's a polynomial (with positive leading coefficient) of odd degree. For such polynomials, one shows that $\lim_{x \to \infty} p(x) = \infty$ and $\lim_{x \to -\infty} p(x) = -\infty$. Then, using the definition of this limit existing, and using the intermediate value theorem, one can show that the range of $p$ is in fact $(-\infty,\infty)$. $\endgroup$ Nov 16, 2021 at 15:58
  • $\begingroup$ @TeresaLisbon Can we conclude if $\lim_{x\to\infty} p(x) = \infty$ and $\lim_{x\to -\infty} p(x) = -\infty$ then $p(x) = (-\infty, \infty)$? How? $\endgroup$
    – user983440
    Nov 16, 2021 at 16:01
  • $\begingroup$ This is probably a duplicate so I'll search for one, but for now : Suppose you want to show that some $r$ belongs in the range of $p$. Using the two limits and appropriately chosen values to fit into the limit definition, you can find $x_1<x_2$ such that $f(x_1) < r < f(x_2)$. Then, IVT applies and for some $c \in [x_1,x_2]$ we must have $f(c) =r$. $\endgroup$ Nov 16, 2021 at 16:04
  • $\begingroup$ You also need the premise that in general, polynomials are continuous functions. $\endgroup$ Nov 16, 2021 at 16:12
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    $\begingroup$ @Yooo Just to commiserate with your latest meta post by showing you my recent comment. [I will disappear this comment later; please do not cite or quote it.] According to the site rules, I cannot reveal whether the culprit is the same entity as said "toxic user". $\endgroup$
    – ryang
    May 18, 2022 at 13:31

1 Answer 1

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Consider any $r \in \mathbb{R}$, then you can show that there exists and $x \in \mathbb{R}$ such that $f(x) = r$. Indeed, $$f(x) = r \Rightarrow f(x)-r = 0$$ but this still is a polynomial of degree $7$. It can be factored over the real numbers into factors of the first and/or second degree. Note that not all factors can be of the second degree (since $f(x)-r$ has degree $7$), so there is at least one factor of degree $1$: $$f(x) - r = (x-a)\cdot q(x)$$ and hence $f(a) = r$.

This shows that the range of $f$ equals $\mathbb{R}$, since $r$ was chosen arbitrarily.

Note: The same argument shows that any polynomial of odd degree has at least one real zero.

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