What is the range and domain of the following: f(x)= x+sqrt(4-x^2)?

Question

I graphed it out, but the graph looks really weird. Please help.

Answer 1

The domain is [-2,2] because outside this interval the radicand is negative.
The function reaches its minimum value of -2 at -2. It reaches its maximum value of 2sqrt(2) at sqrt(2).
Its range is [-2,2sqrt(2)]

Answer 2

The domain is "what values can x have"? We can't have a negative sign under the square root, so we need to find
4 - x^2 >= 0
4 >= x^2
x^2 <= 4
-2 <= x <= 2
These are the values that x must have. So, that's the domain. (x is between -2 and 2, inclusive.)

The range is what possible values of y can we have?

sqrt (4 - x^2) is always going to be between 0 and 2. (because x^2 has to be less than or equal to 4.)

Then we're adding an x to that, where x has to be between -2 and 2 also.

so... the range will be between
-2 + 0
and
2 + 2

so the range will be all values of y between -2 and 4, inclusive.

Answer 3

for the domain your restriction would be that

(4 - x^2) is greater or equal to 0

so

4 - x^2 >= 0

4 >= x^2
or

-2 <= x <= +2

Now your range will be determined by the domain.

Answer 4

To find domain solve

4 - x^2 >= 0


http://www.analyzemath.com/DomainRange/DomainRange.html

Answer 5

concept is that the value inside the radical must not be negative
it must be positive
so that 4-x^2 must be positive or greater than zero
4 - x^2 = > 0
4 => x^2
+- 2 => x
or x <= +- 2 . . . . . . x must less or equal to 2 regardless of sign

Answer 6

f(x) =
x + cos x
px4 + 2

Answer 7

domain [-2,2]