Question

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

Answer 1

Domain=`(-oo,6)` , Range`=RR`. Check below.

Explanation:

`f(x)=ln(6-x)+2`

• For `f` to be defined in `RR` we need:
  `6-x>0` `<=>` `x<6`

Justifications: [ because domain of `y=lnx` is `(0,+oo)` and you have composition of functions `g(x)=lnx` , `h(x)=6-x` so you need `x` `in` `D_h=RR` and `h(x)` `in` `D_g=(0,+oo)` ]

As a result the domain of `f` is `D_f=(-oo,6)`

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'(x)=(ln(6-x)+2)'=1/(6-x)(6-x)'` `=-1/(6-x)=1/(x-6)<0`, if `x<6`
(you can see that by plugging any value `<6` in `f'`)

Therefore `f` is strictly decreasing in `(-oo,6)`

[ Alternative for finding monotony (incase you are not familiar with derivatives):
- Supposed we have `x_1` , `x_2` `in` `(-oo,6)` with `x_1` `<` `x_2`
Then we will have `-x_1>` `-x_2`

`<=>` `6-x_1` `>6-x_2`

`lnx` is increasing in `(0,+oo)` so we can plug `ln` in each side

`<=>` `ln(6-x_1)` `>ln(6-x_2)`

`<=>` `ln(6-x_1)+2` `>ln(6-x_2)+2`

`<=>` `f(x_1)>f(x_2)` `->` `f` strictly decreasing in `(-oo,6)` ]

So we have the following table:

Range will be the ''image'' of the domain,

`f(D_f)=` `f(` `(-oo,6))` `=` `(lim_(xrarr6^(-))f(x),lim_(xrarr-oo)f(x))` `=`

`(-oo,+oo)=RR`

because

• `lim_(xrarr6^(-))f(x)=lim_(xrarr6^(-))` `(ln(6-x)+2)`

Set `6-x=u`

`x->6^-`
`u->0`

`=` `lim_(urarr0)(lnu+2)=-oo+2=-oo`

• `lim_(xrarr-oo)f(x)=lim_(xrarr-oo)(ln(6-x)+2)`

Set `6-x=y`

`x->-oo`
`y->+oo`

`=` `lim_(yrarr+oo)(lny+2)=+oo+2=+oo`

Here is the graph of the function:
`f(x)=ln(6-x)+2` , `x<6`