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Question: Let $$f(x) = \frac{x^6 -1}{3x -1}$$ Prove that the range of $f$ is $\Bbb R$.( Hint: use the Intermediate Value Theorem.)

I thought IVT was meant to show that the function has a root? Please help, I don't know how I can use IVT to prove the range.

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    $\begingroup$ Welcome to Mathematics Stack Exchange. Consider $f(x) $ as $x\to\pm\infty$ $\endgroup$ – J. W. Tanner Sep 16 '19 at 13:20
  • $\begingroup$ Doesn't IVT only work for a closed interval? In this case, f(x) domain is (-infinity, 1/3) U (1/3, infinity). How will this work for an open interval? $\endgroup$ – Amanda Sep 16 '19 at 13:33
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Hint: If $y\in\mathbb R$, then asserting that $y$ belongs to the range of $f$ is the same thing as asserting that the equation $f(x)-y=0$ has a root.

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You have $\lim\limits_{x \to -\infty} f(x)= -\infty$ and $\lim\limits_{x \to \frac{1}{3}^-} f(x)= \infty$.

Moreover $f$ is continuous on the interval $(-\infty, \frac{1}{3})$. Therefore by the IVT, the image of $(-\infty, \frac{1}{3})$ under $f$ is equal to $\mathbb R$. A fortiori, the image of $f$ is equal to $\mathbb R$.

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  • $\begingroup$ This makes sense, but doesn't IVT only work for a closed interval? $\endgroup$ – Amanda Sep 16 '19 at 13:34
  • $\begingroup$ @Armanda You're right that the usual IVT supposes a closed interval. However it is very easy to extend it to an open one. Suppose that $c < y < d$ where $f : (a,b) \to \mathbb R$ with $\lim\limits_{x \to a^+} f(x) = c$ and $\lim\limits_{x \to b^-} f(x) = d$. Take $a < a^\prime < b^\prime < b$ such that $c < f(a^\prime) <y < f(b^\prime) < d$ and apply IVT to the restriction of $f$ to the closed interval $[a^\prime,b^\prime]$. $\endgroup$ – mathcounterexamples.net Sep 16 '19 at 13:44
  • $\begingroup$ Ah, thanks so much, I got it. $\endgroup$ – Amanda Sep 17 '19 at 15:43

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