Alissa Hancock

2022-07-12

Find the domain
a) $f\left(x\right)=\frac{3}{1-4{e}^{3x}}$
b) $g\left(x\right)=\mathrm{ln}\left(5-|x-3|\right)$

Feriheashrz

Expert

a) Given $f\left(x\right)=\frac{3}{1-4{e}^{3x}}$

Value of x will be excluded from the domain as the restriction of function denominator
$\left(1-4{e}^{3x}\right)\ne 0$

If $\left(1-4{e}^{3x}\right)=0$
$1=4{e}^{3x}⇒\frac{1}{4}={e}^{3x}$
$\mathrm{log}\left(\frac{1}{4}\right)=\mathrm{log}\left({e}^{3x}\right)⇒\mathrm{log}\left(\frac{1}{4}\right)=3x\mathrm{log}\left(e\right)$
$x=\frac{\mathrm{log}\left(\frac{1}{4}\right)}{3}=\frac{1}{3}\mathrm{log}\left(\frac{1}{4}\right)$

Thus, the domain of the function is $\left(-\mathrm{\infty },\frac{1}{3}\mathrm{log}\left(\frac{1}{4}\right)\right)\cup \left(\frac{1}{3}\mathrm{log}\left(\frac{1}{4}\right),\mathrm{\infty }\right)$

b) Given $g\left(x\right)=\mathrm{ln}\left(5-|x-3|\right)$
As logarithm is only defined for positive numbers
$\left(5-|x-3|\right)>0$
$5>|x-3|$

And we get two conditions:
if $x>3$
$g\left(x\right)=\mathrm{ln}\left(5-x+3\right)$
$g\left(x\right)=\mathrm{ln}\left(8-x\right)$

And $8-x>0$
$x>8$

If $x<3$
$g\left(x\right)=\mathrm{ln}\left(5+x-3\right)$
$g\left(x\right)=\mathrm{ln}\left(2+x\right)$
And $2+x>0$
$x>-2$

Thus, the domain of the function is $\left(-2,8\right)$

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