First you need to be sure to avoid xx values that makes the argument of your square root negative (you cannot find a real solution of a negative square root). So you set:
4-x^2>=04−x2≥0 and:
(2-x)(2+x)>=0(2−x)(2+x)≥0
so:
x>=-2x≥−2
x<=2x≤2
and -2<=x<=2−2≤x≤2
Outside this interval your function does not exist.
So, now we choose values for xx and evaluate the correspondent yy:
x=2 => y=0x=2⇒y=0
x=-2 => y=0x=−2⇒y=0
x=0 => y=0x=0⇒y=0
We have a graph that passes through y=0y=0 3 times; probably our function changes quadrant in doing so.
For xx approaching -2−2 our function gives negative values (for example when x=-1.9=> y=-1.2x=−1.9⇒y=−1.2) and when approaching 22 it has positive values (for example when x=1.9=> y=1.2x=1.9⇒y=1.2). The graph of our function starts from x=-2x=−2 going downwards, rises passing through x=0x=0 and then bends again to end up in x=2x=2. I must have a minimum and a maximum somewhere.
I evaluate the derivative of the function and set it equal to zero to find these points:
f'(x)=sqrt(4-x^2)-x^2/(sqrt(4-x^2))
setting it equal to zero and manipulating it I got:
4-2x^2=0 which gives x=+-sqrt(2)
so you get:
x=-sqrt(2) => y=-2 minimum and
x=sqrt(2) => y=2 maximum
Finally:
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