Question

domain and range of √9-x²

Answer 1

Answer:

f(x)=$\sqrt{9-x^2}$

Domain of the equation,

9-$x^{2}$≥0  [Since,for real numbers sq rt is always a positive number],

$x^{2}$-9≤0

(x+3)(x-3)=0

∴Domain⇒-3≤x≤3

f(x)=$\sqrt{9-x^2}$ is a semicircle having radius 3 and origin (0,0), [y=$\sqrt{r^2-x^2}$ represents a semicircle,where r is the radius],

∵The range is the set of possible output values, which are shown on the y-axis.

And,since the maximum value of the semicircle on y-axis is 3 as the origin of the circle is (0,0) and radius is 3,

∴Range=[0,3]

⇒Pls refer the attachment of the graph given below,

Hope it helps you.

Answer 2

Answer:

values of x will be restricted such that ( 9 - x^2 ) > 0

9 - x^2 = ( 3 + x ) ( 3 - x )

this is parabola of inverted U shape, whose value is +ve in between roots.

so ( 9 - x^2 ) > 0 for x in between -3 & 3

so domain is x = [ -3 , 3 ]

range = ( 0 , max of f(x) )

f(x) is maximum at mid point of roots, i.e. f max = f(0) = 3

so range = ( 0 , 3 )