If $f(x)$ is the square root of the number that is $2$ more than $x,$ what is the value of $f(7) - f(-1)$?

Question

If $f(x)$ is defined for all $x>-2$ as the square root of the number that is $2$ more than $x$, what is the value of $f(7) - f(-1)$?

My attempted solution:

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

Now, $f(7) = \sqrt{7 + 2} = \pm3$

and,

$f(-1) = \sqrt{-1 + 2} = \pm1$.

So, $f(7) - f(-1) = 2, 4, -4, -2$.

Given answer is, $2$ only.

Why?

Answer 1

The given answer is taking what is referred to as the principal square root, rather like $x=\frac{\pi}{6}^c$ or $x=30^0$ is the principal root of $\sin(x)=\frac{1}{2}$, there are many other solutions, but this is the one that is most commonly used, and given by calculatorss.

In this question, it is standard to use the principal square root for equations like these. Hence we get $\sqrt{9}=3, \sqrt{1}=1$, thus $f(7)-f(-1)=3-1=2$

Answer 2

The square root of $1$ is $1$, not $±1$ and so is the square root of $9$.

So the answer will be $3-1 = 2$

Answer 3

It is a convention that for $a>0$, if $$b=\sqrt a$$, then $b>0$.

In some articles, $b$ is called the principal value of square root, distinguishing it from the other possible value, $-b$.