Find domain and range of f(x) = √(4-x²)

Question

Find the domain and range of $f(x) = \sqrt{4 - x^2}$ .

Solution — Step by Step

For $f(x) = \sqrt{4 - x^2}$ to be defined over real numbers, the expression under the square root must be non-negative:

4 - x^2 \geq 0

This is the only restriction (square roots of negative numbers are not real).

4 - x^2 \geq 0

x^2 \leq 4

|x| \leq 2

-2 \leq x \leq 2

Domain $= [-2, 2]$

Now we need all possible values $f(x)$ can take when $x \in [-2, 2]$ .

Since $f(x) = \sqrt{4 - x^2}$ , and square root is always $\geq 0$ , the range contains only non-negative values.

When is $f(x)$ minimum? When $4 - x^2$ is minimum, i.e., when $x^2$ is maximum. Maximum $x^2$ on $[-2,2]$ is 4 (at $x = \pm 2$ ).

f(\pm 2) = \sqrt{4 - 4} = \sqrt{0} = 0

When is $f(x)$ maximum? When $4 - x^2$ is maximum, i.e., when $x^2$ is minimum. Minimum $x^2 = 0$ (at $x = 0$ ).

f(0) = \sqrt{4 - 0} = \sqrt{4} = 2

Since $f(x)$ is continuous on $[-2, 2]$ and takes all values from 0 to 2 (by the intermediate value theorem), the range is:

Range $= [0, 2]$

Final Answer:

Domain: $[-2, 2]$
Range: $[0, 2]$

Why This Works

The function $f(x) = \sqrt{4 - x^2}$ is the upper semicircle of the circle $x^2 + y^2 = 4$ (radius 2, centred at origin). The square root gives only the positive $y$ values.

Seeing this geometrically makes the answer obvious: the semicircle’s $x$ -values run from $-2$ to $2$ (domain), and the $y$ -values run from 0 (at the endpoints) to 2 (at the top, the apex). So domain $= [-2,2]$ and range $= [0,2]$ .

This geometric interpretation is a powerful cross-check. Any time you see $\sqrt{r^2 - x^2}$ , think: semicircle of radius $r$ .

Alternative Method — Substitution

Let $y = \sqrt{4 - x^2}$ . Then $y \geq 0$ and $y^2 = 4 - x^2$ , so $x^2 = 4 - y^2$ .

For real $x$ : $4 - y^2 \geq 0 \Rightarrow y^2 \leq 4 \Rightarrow 0 \leq y \leq 2$ (since $y \geq 0$ ).

This confirms range $= [0, 2]$ directly.

Common Mistake

Writing domain as $(-2, 2)$ — open interval — instead of $[-2, 2]$ — closed interval. At $x = \pm 2$ , the expression under the square root equals zero, and $\sqrt{0} = 0$ is perfectly valid. Zero under the square root is not a problem — only negative values are. So $x = \pm 2$ are included in the domain.

For domain questions: identify restrictions first. Square roots require non-negative inside. Logarithms require positive inside. Denominators require non-zero. Then solve the corresponding inequality. The solution set is your domain.

Question

Solution — Step by Step

Why This Works

Alternative Method — Substitution

Common Mistake

Try These Next