Why Are Negative Values Excluded in the Range of These Square Root Functions?

[Michael_Light]

Homework Statement

Find the range of each of the following functions. All the functions are defined for the largest possible domain of values of x.

a) f(x) = √(4-x^2) b) f(x) = √(4-x)

Homework Equations

The Attempt at a Solution

The answers given are a) 0 ≤ f(x) ≤ 2 b) f(x) ≥ 0 . But my answers are a) -2≤ f(x)≤2 b) All real numbers . Can anyone explain what i had done wrong? Why negative numbers are excluded?

 

[eumyang]

Because you wouldn't have a function otherwise. Consider the basic square root function
$$f(x) = \sqrt{x}$$

You're probably thinking that there are two square roots of a number,
$$\pm\sqrt{x}$$

However, in the function
$$f(x) = \sqrt{x}$$
if we allow both positive and negative values, you would end up with a single x-value paired with two function values (like (16, 4) and (16, -4)). That's not allowed in functions.

Your original problem works the same way. The negative values will not be in the range, because otherwise you wouldn't have functions anymore.

 

[symbolipoint]

Notice in exercise #b, if x>4, then the function has no Real value. Also, the square root will not be less than 0, meaning the function will be in range of greater or equal to 0than 0 but not less than 0.

 

[HallsofIvy]

Homework Statement

Find the range of each of the following functions. All the functions are defined for the largest possible domain of values of x.

a) f(x) = √(4-x^2) b) f(x) = √(4-x)

Homework Equations

The Attempt at a Solution

The answers given are a) 0 ≤ f(x) ≤ 2 b) f(x) ≥ 0 . But my answers are a) -2≤ f(x)≤2 b) All real numbers . Can anyone explain what i had done wrong? Why negative numbers are excluded?

Because, as eumyang said, $\sqrt{4- x^2}$ is defined as the positive number such that its square is $4- x^2$. Similarly, $\sqrt{4- x}$ is defined as the positive number whose square is 4- x.