# Let f ( x ) = x 6 </msup> &#x22

Let
$f\left(x\right)=\frac{{x}^{6}-1}{3x-1}$
Prove that the range of $f$ is $\mathbb{R}$.( 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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Dobermann82
You have $\underset{x\to -\mathrm{\infty }}{lim}f\left(x\right)=-\mathrm{\infty }$ and $\underset{x\to {\frac{1}{3}}^{-}}{lim}f\left(x\right)=\mathrm{\infty }$.

Moreover $f$ is continuous on the interval $\left(-\mathrm{\infty },\frac{1}{3}\right)$. Therefore by the IVT, the image of $\left(-\mathrm{\infty },\frac{1}{3}\right)$ under $f$ is equal to $\mathbb{R}$. A fortiori, the image of $f$ is equal to $\mathbb{R}$.
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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 $f\left(x\right)-y=0$ has a root.