Trouble understanding range values

[phospho]

For example: f(x) = 2 - √(x+9), x ≥ -9

I'm asked to find the inverse and the domain of the inverse which is the range of f(x). I've found the inverse but for the domain, I put x ε ℝ however the answer is x≥2

I can see how they got the answer but why is my answer incorrect. This is how I'm seeing it:

for example: x = - 8 => f(x) = 2 - ±√(-8+9) leading to f(x) = 1, 3: but they seem to be taking the positive root only. Why?

 

[Ray Vickson]

For example: f(x) = 2 - √(x+9), x ≥ -9

I'm asked to find the inverse and the domain of the inverse which is the range of f(x). I've found the inverse but for the domain, I put x ε ℝ however the answer is x≥2

I can see how they got the answer but why is my answer incorrect. This is how I'm seeing it:

for example: x = - 8 => f(x) = 2 - ±√(-8+9) leading to f(x) = 1, 3: but they seem to be taking the positive root only. Why?

By definition, the "square-root" function is always ≥ 0. That is, the two roots of an equation of the form $u = v^2$ are $v = \pm \sqrt{u}$; $\sqrt{u}$ itself is never < 0 (for non-negative real values of u).

In your example, the values of $y = 2 - \sqrt{x+9}$ are always ≤ 2, and for any such y the corresponding x is obtained from
$$\sqrt{x+9} = 2-y \: \Longrightarrow x+9 = (2-y)^2,$$
hence
$$f^{-1}(y) = (2-y)^2 - 9, \: y \leq 2.$$

Of course, you can re-write this by using the symbol x instead of the symbol y; the domain would then be x ≤ 2 (NOT x ≥ 2).

 

[phospho]

By definition, the "square-root" function is always ≥ 0. That is, the two roots of an equation of the form $u = v^2$ are $v = \pm \sqrt{u}$; $\sqrt{u}$ itself is never < 0 (for non-negative real values of u).

In your example, the values of $y = 2 - \sqrt{x+9}$ are always ≤ 2, and for any such y the corresponding x is obtained from
$$\sqrt{x+9} = 2-y \: \Longrightarrow x+9 = (2-y)^2,$$
hence
$$f^{-1}(y) = (2-y)^2 - 9, \: y \leq 2.$$

Of course, you can re-write this by using the symbol x instead of the symbol y; the domain would then be x ≤ 2 (NOT x ≥ 2).

I see, thanks.