SOLUTION: Hello. I need to find the natural domain of f(x)=sqrt x(16-x^2) Note: Just to be clear, "x(16-x^2)" is UNDER the square root sign. Thanks.

Algebra ->  Functions -> SOLUTION: Hello. I need to find the natural domain of f(x)=sqrt x(16-x^2) Note: Just to be clear, "x(16-x^2)" is UNDER the square root sign. Thanks.       Log On


   

Question 795636: Hello.
I need to find the natural domain of f(x)=sqrt x(16-x^2)
Note: Just to be clear, "x(16-x^2)" is UNDER the square root sign.
Thanks.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29+=+sqrt%28+x%2A%28+16+-+x%5E2+%29+%29+
If +x+=+0+, the function exists- it is +0+
If +x+=+4+ or
if +x+=+-4+, +f%28x%29+=+0+ also
----------------------------
If +x+%3E+4+, +f%28x%29+ is the square root
of a negative number, so the function
does not exist if +x+%3E4+
-----------------------
If +x+%3C+-4+ then +sqrt%28+x%2A%28+16+-+x%5E2+%29+%29+ is
the square root of a negative times a negative,
so the function exists
-----------------------
If +-4+%3C+x+%3C+0+, f(x) does not exist because you
have the square root of a negative times a positive
which is a negative
------------------------
If +0+%3C+x+%3C+4+, f(x) exists because you have
the square root of a positive times a positive
------------------------
Putting this together, the domain is:
+x+%3C+=+-4+
+0+%3C+=+x+%3C=+4+
----------------
Here's the plot:
+graph%28+400%2C+400%2C+-6%2C+6%2C+-8%2C+8%2C+sqrt%28+x%2A%28+16+-+x%5E2+%29%29++%29+