question about continuous functions?

Question

we would like the function g(x)= square root of ( 16 - x^2 / 4 - x)
-4 < or = x < or = 4

to be continuous over the interval (-4,4). The value that would have to be assigned to g(4) to accomplish this is:
a) 0 b) 2 c) 2 * sqrt 2 d) 4 e) 4 * sqrt 2

and why.


thank you soooooo much

Answer 1

You mean over the interval [-4,4], right? If so, just find the limit of your function as x goes to 4. You can do this by taking the limit under the square root and doing a cancellation. By the way DON'T FORGET PARENTHESES! You meant to write (16-x^2)/(4-x).
Since that is the same as 4+x, the limit is easy to take. The value of the limit is the value the function needs to take when x=4.

Answer 2

16-x^2/4 -x=0
64-x^2-4x=0

quadratic formula:
x=[ 4+-sqrt(16-4(-1)64)] /2(-1)
x= [4 +- sqrt(272)]/-2
x=-10.246 or x=6.246
that means that the function g(x) is continuos, since
16 - x^2 / 4 - x>0 if -4 < = x < = 4

now g(4)=sqrt(16-16/4 -4)
=sqrt(16-4-4)
=sqrt(8)
=2sqrt(2)

Answer 3

To solve this problem start by factoring the numerator:
(4-x)(4+x)/(4-x) Now just cancel
4+x But remember that x still can not equal 4
When x=4, 4+x=8
therefore, g(4)=8 will make this function continuous

Answer 4

Well there is the hole at 4. The answer is C) 2 sqrt2. If you graph the two functions they intersect exaclty at 4. I am not sure that that is the answer your teacher will want tho.