Find Dom(f) and Range(f) of g(x)=(In(x-5)/In(x-5))+sqrt(10-x)

[philippe311]

Homework Statement

g(x)=(In(x-5)/In(x-5))+sqrt(10-x). Find Dom(f) and Range(f).
and sqrt(10-x) is decreasing.

The Attempt at a Solution

x-5=0
x=5.
10-x>=0
10>=x.
so The dom(f), I think, is (-infinity,5)U(5,10].
now the range, the equation should look like this after cancelling.
1+sqrt(10-x).
Is it right that I just plug in x values like:10,5,...
and I tried to plug in some and I got this
(1,1+sqrt(5))U(1+sqrt(5),infinity).
Is that right? Or am I missing something?

 

[Mark44]

Homework Statement

g(x)=(In(x-5)/In(x-5))+sqrt(10-x). Find Dom(f) and Range(f).
and sqrt(10-x) is decreasing.

The Attempt at a Solution

x-5=0
x=5.
10-x>=0
10>=x.
so The dom(f), I think, is (-infinity,5)U(5,10].
now the range, the equation should look like this after cancelling.
1+sqrt(10-x).
Is it right that I just plug in x values like:10,5,...
and I tried to plug in some and I got this
(1,1+sqrt(5))U(1+sqrt(5),infinity).
Is that right? Or am I missing something?

First off, there is no In function, just as there is no Iog function. The names of these function are ln and log, respectively, for the natural logarithm and common logarithm functions.

Your domain is not correct. For ln(x - 5) to be defined, x > 5, which you have. For the square root, x <= 10, which is what you have. For ln(x-5)/ln(x - 5) to be defined, however, you can't divide by zero. What value of x makes ln(x - 5) = 0?

As already mentioned in your problem, sqrt(10 - x) is decreasing, so 1 + sqrt(10 - x) will be decreasing as well. The restricted domain will affect the graph of 1 + sqrt(10 - x), so will affect the range of your function f.

 

[philippe311]

Thanks for explaining.