How do you find the domain and range of $g (x) = \frac{1}{I n x}$ $g(x) = 1/(In x)$ ?

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Answer 1 · 2017-07-03T10:50:49

Domain: $(0, 1) \cup (1, + \infty)$ $(0, 1)uu(1, +oo)$
Range: $(- \infty, + \infty)$ $(-oo, +oo)$

$g (x) = \frac{1}{ln x}$ $g(x) = 1/lnx$

$ln x$ $lnx$ is defined for $x > 0$ $x>0$

$ln x = 0$ $lnx = 0$ for $x = 1$ $x=1$ $\to g (x)$ $-> g(x)$ is not defined for $x = 1$ $x=1$

Hence, $g (x)$ $g(x)$ is defined for $x > 0, x \neq 1$ $x>0, x!=1$

$:.$ the domain of $g(x)$ is $(0, 1)uu(1, +oo)$

Now consider:

$lim_"(x->1) -"g(x) = -oo$

$lim_"(x->1) +"g(x) = +oo$

Hence, the range of $g(x)$ is $(-oo, +oo)$

We can see these results fron the graph of $g(x)$
below.

graph{1/lnx [-10, 10, -5.04, 4.96]}

How do you find the domain and range of g(x) = 1/(In x)g(x)=1Inxg(x) = 1/(In x)?