Question

Find the domain and range of the following real function:

1. f(x) = |x|
2. f(x)=$\sqrt {9-x^2}$

Answer 1

(i) f(x) = |x|, x ∈ R
We know that:

$|x|= \begin{cases} x, & \quad \text{if } x≥0\\\ -x, & \quad \text{if } n<0 \end{cases}$

$∴ f(x)=-|x|= \begin{cases} -x, & \quad \text{if } x≥0\\\ x, & \quad \text{if } n<0 \end{cases}$

Since f(x) is defined for x ∈ R, the domain of f is R.
It can be observed that the range of f(x) = |x| is all real numbers except positive real numbers.
∴ The range of f is (∞, 0].

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(ii) f(x)=$\sqrt {9-x^2}$
Since $\sqrt {9-x^2}$ is defined for all real numbers that are greater than or equal to 3 and less than or equal to 3, the domain of f(x) is {x : 3 ≤ x ≤ 3} or [3, 3].
For any value of x such that 3 ≤ x≤ 3, the value of f(x) will lie between 0 and 3.
∴ The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].