Question

Find the range of the real valued function: f(x) = $\sqrt{9 + x^{2}}$

Answer 1

The given function is

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

It is valid for all values of x because there is x².

And value of x² is always positive. And there is no -ve sign in the function. So all the outputs will be +ve.

The minimum value of the function $f(x)=\sqrt{x^2+9}$ is "3" for "x=0"

$f(x)=\sqrt{x^2+9}=\sqrt{0^2+9}=\sqrt{9}=3$

So

Domain = Set of real numbers

Range = [ 0 , ∞ ) = All non-negative Real Numbers

Answer 2

Answer:

range (f)=[0,∞]

Step-by-step explanation:

To find the range of the real valued function:

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

let

y= f(x)

$y=\sqrt{9+x^{2}} \\\\y^{2} =9+x^{2} \\\\x^{2}=y^{2}-9\\\\x=\sqrt{y^{2}-9}$

it is clear that x will take real value only if

$\sqrt{y^{2}-9}$ ≥0

y²-9≥0

(y-3)(y+3)≥0

=> -3 ≤ y ≤3

=> y∈ [-3,3]

∵ y= √(9+x²)≥0 for all R

so range (f)=[0,∞]