Question

Find the domain and range:

f(x)=sqrt(32+4x^2)

Calculate the exact value(s) of x when f(x)=6.

Answer 1

that will give you your domain

Answer 2

solve 32+4x^2>or equal to 0

Answer 3

I gathered no solution.

Answer 4

4x^2>=-32
x^2>=-8
are you sure it's no solutions?

Answer 5

Isnt domain all real numbers?

Answer 6

The domain is all real numbers. The range is f(x) is greater than sqrt(32)

Answer 7

http://www.wolframalpha.com/input/?i=f%28x%29%3Dsqrt%2832%2B4x%5E2%29
4x^2=4
X^2=1
X=+-1

Answer 8

forgot to put beginning
sqrt(32+4x^2)=6
32+4x^2=36

Answer 9

wrt domain you only have to find any values of x that would make 32+4x^2 negative.... and there are none! the only way that expression could even be 0 is if the squared term equals -32. no calculations necessary

Answer 10

the system says incorrect...no luck with the domain. :\

Answer 11

domain should be all real numbers because you can never have a zero under the square root bcz of the square on x

Answer 12

also u cant have ive under the square root bcz of x^2

Answer 13

-ive*