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I have the multivariable function

$$\log(y^2+4x^2-4)$$

and I have found the maximal domain to be

$$x^2+{y^2\over4}>1$$

However I have no idea how to find the range of this function, any insight would be greatly appreciated

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  • $\begingroup$ Take $y=0.$ Then you can study the range of the function of one variable. Once you got it you will know the range of the multivariable function. $\endgroup$ – mfl Mar 21 '15 at 1:12
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Hint. We have $$ \ln : (0,\infty) \longrightarrow \mathbb{R} $$ and $$ \ln : (1,\infty) \longrightarrow \mathbb{R}^+-\left\{0\right\}. $$

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  • $\begingroup$ So if I were to let y = 0, $$z=log(4x^2-4)$$ Which therefore the range = (1,inf) Is this correct? $\endgroup$ – user145608 Mar 21 '15 at 2:18
  • $\begingroup$ @user145608 Looking at my second proposition above, you may see that the answer to your question is $\mathbb{R}^+-\left\{0\right\}$, since the argument of your function, that is $x^2+{y^2\over4}$, is greater than $1$. Thanks! $\endgroup$ – Olivier Oloa Mar 21 '15 at 6:24
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You have already found out that the domain $D$ of $f$ is the open exterior of the ellipse $x^2+{y^2\over4}=1$. The set $D$ contains all points of the form $(x,0)$ with $x>1$, and on these points $f$ assumes the values $$\phi(x):=\log(4x^2-4)=\log 4+\log(x+1)+\log(x-1)\ .$$ From $$\lim_{x\to1+}\phi(x)=-\infty,\qquad\lim_{x\to\infty}\phi(x)=\infty$$ it then follows that $f$ assumes arbitrary real values; whence the range of $f$ is ${\mathbb R}$.

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