# How do you find the domain and range of y= Ln(6-x) +2?

Jan 11, 2018

Domain=$\left(- \infty , 6\right)$ , Range$= \mathbb{R}$. Check below.

#### Explanation:

$f \left(x\right) = \ln \left(6 - x\right) + 2$

• For $f$ to be defined in $\mathbb{R}$ we need:
$6 - x > 0$ $\iff$ $x < 6$

Justifications: [ because domain of $y = \ln x$ is $\left(0 , + \infty\right)$ and you have composition of functions $g \left(x\right) = \ln x$ , $h \left(x\right) = 6 - x$ so you need $x$$\in$${D}_{h} = \mathbb{R}$ and $h \left(x\right)$$\in$${D}_{g} = \left(0 , + \infty\right)$ ]

As a result the domain of $f$ is ${D}_{f} = \left(- \infty , 6\right)$

For the range i will work with monotony and continuity of the function.

$f$ is continuous/differentiable in ${D}_{f}$ as a function of the compositions mentioned above.

$f ' \left(x\right) = \left(\ln \left(6 - x\right) + 2\right) ' = \frac{1}{6 - x} \left(6 - x\right) '$ $= - \frac{1}{6 - x} = \frac{1}{x - 6} < 0$, if $x < 6$
(you can see that by plugging any value $< 6$ in $f '$)

Therefore $f$ is strictly decreasing in $\left(- \infty , 6\right)$

[ Alternative for finding monotony (incase you are not familiar with derivatives):
- Supposed we have ${x}_{1}$ , ${x}_{2}$ $\in$$\left(- \infty , 6\right)$ with ${x}_{1}$$<$${x}_{2}$
Then we will have $- {x}_{1} >$$- {x}_{2}$

$\iff$ $6 - {x}_{1}$$> 6 - {x}_{2}$

$\ln x$ is increasing in $\left(0 , + \infty\right)$ so we can plug $\ln$ in each side $\iff$ $\ln \left(6 - {x}_{1}\right)$$> \ln \left(6 - {x}_{2}\right)$

$\iff$ $\ln \left(6 - {x}_{1}\right) + 2$$> \ln \left(6 - {x}_{2}\right) + 2$

$\iff$ $f \left({x}_{1}\right) > f \left({x}_{2}\right)$ $\to$ $f$ strictly decreasing in $\left(- \infty , 6\right)$ ]

So we have the following table: Range will be the ''image'' of the domain,

$f \left({D}_{f}\right) =$f((-oo,6)) $=$ $\left({\lim}_{x \rightarrow {6}^{-}} f \left(x\right) , {\lim}_{x \rightarrow - \infty} f \left(x\right)\right)$ $=$

$\left(- \infty , + \infty\right) = \mathbb{R}$

because

• ${\lim}_{x \rightarrow {6}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {6}^{-}}$$\left(\ln \left(6 - x\right) + 2\right)$

Set $6 - x = u$

$x \to {6}^{-}$
$u \to 0$

$=$ ${\lim}_{u \rightarrow 0} \left(\ln u + 2\right) = - \infty + 2 = - \infty$

• ${\lim}_{x \rightarrow - \infty} f \left(x\right) = {\lim}_{x \rightarrow - \infty} \left(\ln \left(6 - x\right) + 2\right)$

Set $6 - x = y$

$x \to - \infty$
$y \to + \infty$

$=$ ${\lim}_{y \rightarrow + \infty} \left(\ln y + 2\right) = + \infty + 2 = + \infty$

Here is the graph of the function:
$f \left(x\right) = \ln \left(6 - x\right) + 2$ , $x < 6$ 