##### main algebraic solution AND CHECK!!

label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

May 27th, 2015

First we need to determine when f(x) is invalid. In this case we look at the roots which are x-6 and x+2 so this means f(x) domain cannot include x=6 or x=-2

then we look at g(x) and its the sqrt of 4-x and a sqrt must be equal or greater to 0. so 4-x>=0 x<=4 i switch signs because of the negative

The reason i did not do the numerator of f(x) since nothing there will cause it to be undefined. So now that we have the domain of our f(x) and g(x) we now must simplify it and see what x=values are allowed and what arent

So now we combine our domains and since we must be less than or equal to 4 or domain is anything less than or equal to 4 that does not include x=-2, x=6 would cause our sqrt to be a negative and it is not allowed. As for x=-2 it would result in the f(x) being undefined and it would result in f(x)/g(x) being undefined

Interval notation

(-inf,-2),(-2,4] this shows that x can be anything negative infinite to -2 but not including -2 and then anything between -2 and 4 including 4. Plugin in the numbers to verify if you get undefined then it is not in the domain. Hope this helps

May 27th, 2015

at -2 you would get f(x) denominator equalling 0 since 4--8-12 =0 and at 5 sqrt(4-5)=-1 and you cannot sqrt a negative

May 27th, 2015

...
May 27th, 2015
...
May 27th, 2015
Oct 19th, 2017
check_circle