Question

(x) = \sqrt{9 - {x}^{2} } " alt="f(x) = \sqrt{9 - {x}^{2} } " align="absmiddle" class="latex-formula">
I have a doubt. I know that the domain will be [-3, 3] because this is a real function. But what will be the range. Everywhere it is given as [0, 3]. But shouldn't it be [-3, 3].

For example let x = 0, then f(x) = √9, and it can be either -3 or 3. Why are we considering just the positive value?

Please explain.​

Answer 1

Answer:

range of f is [0,3]

Step-by-step explanation:

The "square root" symbol, $\sqrt{\text{ \quad }}$, is a function.  As a result, it only has a single output because functions must only have a single output for each input.

Thinking about it graphically, if the square root function did give both a positive and a negative result, the function "f" would not pass the vertical line test and it would not be a function.

When solving an equation like $x^2=25$, to solve, we must apply the square root property.  The square root property says that to find both solutions, one must look at both the positive and the negative of the square root.  So, to solve:

$x^2=25$

Apply square root property...

$\sqrt{x^2} = \pm \sqrt{25}$

$x=\sqrt{25}$  or  $x=-\sqrt{25}$

$x=5$  or  $x=-5$

In this case, because the square root function itself only outputs non-negative results (so, including zero, as you already identified), the range will only be [0,3].