find the range of:
f(x)=sqrt(16-x2)
we get the domain as [-4,4] but we only take [0,4] as the range. can someone tell me why dont we take the negative values
1.Because y=f(x)=Sq.Rt( 16-X 2 ) is actually a circle and the portion of the circle in the range -4 to 0 is not to be considered as this will violate the definition of a function.
2. It fails the vertical line test of a function that is for the same value of independent variable (x) you will get two vlue of f(x)
3. Cosnsider y 2= sq.rt (16 - x 2 ) here for the same value of x in the Domain -4 to +4 say x=2 you will get y =sq.rt of 12 =( + - ) 3.464 . now the value
-3.464 falls in the portion of the circle below x axis and hence this cannot be considered as a valid point for the function in the domain of definition -4 to +4.
I hope I have made it abudantanly clear your doubt.
If the question is
\(y=\sqrt{16-x^{2}}\), it means it is
\(y=+\sqrt{16-x^{2}}\). Hence, no question of taking the negative value of y as the answer.