# How to find range of $f(x) = \frac{x^7 + x^5 + x^3 - x^2 - 1}{6}$

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

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?

• 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 $(-\infty,\infty)$. Nov 16, 2021 at 15:58
• @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?
– user983440
Nov 16, 2021 at 16:01
• 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 $f(c) =r$. Nov 16, 2021 at 16:04
• You also need the premise that in general, polynomials are continuous functions. Nov 16, 2021 at 16:12
• @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 [I will disappear this comment later; please do not cite or quote it.]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". May 18, 2022 at 13:31

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.