Question

find the domain range of f(x)=1/√9-x^2​

Answer 1

Domain: -3 < x<3 ,
Range:. y>1/3
I hope you will get it easily.

Answer 2

Answer:

Domain: (-3,3)

Range: {1/3, infinity)

Step-by-step explanation:

All those values of X that are specifically real values of X that it can accept so that the value of the function comes out to be a precisely defined real number are referred to as the domain:

Domain:

$F(x)= 1\sqrt{(9-x^2)$

$9-x^2 > 0$

$x^2 < 9$

X belongs to $(-3,3)$

Range:

$1/\sqrt{(9-x^2)$

$x^2 min =0$

$\sqrt{(9-x^2)} = 3 = max$

$1/\sqrt{(9-x^2 )} = 1/3 = min$

$x^2 = 9= max$

$1/\sqrt{(9- x^2 )} = infinity = max$

Range = $[\frac{1}{3}, infinity)$

#SPJ2