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Problem: Find the domain and range of the function $f(x)=\log \sqrt{4-x^2}$

My attempt:

For domain

$y=f(x)=\log \sqrt{4-x^2}$

$\implies y=\log \sqrt{4-x^2}$

Since logarithm of only positive numbers is possible,

$4-x^2>0$

$\implies x^2-4<0$

$\implies(x+2)(x-2)<0$

$\implies -2<x<2$

Therefore $D_f=(-2,2)$

For range

$y=f(x)=\log \sqrt{4-x^2}$

$y=\log \sqrt{4-x^2}$

$y=\frac{1}{2}\log (4-x^2)$

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

$e^{2y}=4-x^2$

$x=\sqrt{4-e^{2y}}$

Since the exponent can take any real value, $R_f=(-\infty,+\infty)$

My problem: The answer given in my book for $D_f$ matches with my $D_f$ but $R_f=(-\infty,\log 2]$ according to my book. Where have i gone wrong?

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$\sqrt {4-x^2}$ cannot be greater than $\sqrt 4=2$, so $\log \sqrt{4-x^2}$ cannot be any greater than $\log \sqrt 4=\log 2$. It achieves that value when $x=0.$ As $\sqrt{4-x^2}$ can achieve $0$, you can have $\log \sqrt{4-x^2}$ as negative as you want, so the range is $(-\infty, \log 2]$

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  • $\begingroup$ Where have i gone wrong? $\endgroup$ – MrAP Jul 9 '17 at 21:22
  • $\begingroup$ The whole approach of solving for $x$ is misguided. It will fail for functions that are not bijections. You have a continuous function so only need to find the maximum and minimum. Taking the square root is dangerous. Finally, though $e^{2y}$ alone can take any positive value, if it gets greater than $4$ the square root argument becomes negative. $\endgroup$ – Ross Millikan Jul 9 '17 at 21:30
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Exponent can't take any real value.

For $x \in \mathbb R$, we must have $$x=\sqrt{4-e^{2y}} \in \mathbb R \implies 4-e^{2y} \ge 0 \implies y \in (-\infty,\ln 2)$$

This gives you the correct answer.

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  • $\begingroup$ Why "Exponent can't take any real value."? $\endgroup$ – MrAP Jul 9 '17 at 18:50
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Yes, the exponent can take any value, but if $y\geqslant\log2$, then $e^{2y}\geqslant4$, and therefore, $4-e^{2y}\leqslant0$.

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