# What is the range of the function h(x) = ln(x+6)?

Nov 16, 2017

Answer: Using Monotony/continuity & Domain: $h \left(D h\right) = R$

#### Explanation:

$h \left(x\right) = \ln \left(x + 6\right)$ , $x >$$- 6$

$D h = \left(- 6 , + \infty\right)$

$h ' \left(x\right) = \frac{1}{x + 6}$$\left(x + 6\right) '$$= \frac{1}{x + 6}$ $> 0$, $x > - 6$
So that means that $h$ is strictly increasing ↑ in $\left(- 6 , + \infty\right)$

$h$ is obviously continuous in $\left(- 6 , + \infty\right)$ as composition of ${h}_{1}$(x)=x+6 & ${h}_{2}$(x) = $\ln x$

h(Dh)=h((-6,+oo))= (${\lim}_{x \rightarrow - 6} h \left(x\right)$,lim_(xrarr+oo)h(x)) $= \left(- \infty , + \infty\right)$$= R$

because ⋅${\lim}_{x \rightarrow - 6} h \left(x\right)$= ${\lim}_{x \rightarrow - 6} \ln \left(x + 6\right)$

$x + 6 = y$
$x \rightarrow - 6$
$y \rightarrow 0$

$= {\lim}_{y \rightarrow 0} \ln y$ $= - \infty$

⋅${\lim}_{x \rightarrow + \infty} h \left(x\right)$=${\lim}_{x \rightarrow + \infty} \ln \left(x + 6\right)$$= + \infty$

Note: you can also show this using the reverse ${h}^{-} 1$ function. (y=ln(x+6)=>......)