# Determining range using Intermediate Value Theorem

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.

• Welcome to Mathematics Stack Exchange. Consider $f(x)$ as $x\to\pm\infty$ – J. W. Tanner Sep 16 '19 at 13:20
• 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? – Amanda Sep 16 '19 at 13:33

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.
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 $$\mathbb R$$.
• @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 $[a^\prime,b^\prime]$. – mathcounterexamples.net Sep 16 '19 at 13:44