Q.Find the range of each of the following functions.

Concept: Range of a Function

The range is the set of all possible output values $f(x)$ as $x$ varies over the given domain.

(i) $f(x) = 2 - 3x,\ x > 0$

As $x$ increases from just above 0 to $\infty$, $f(x)$ decreases from just below 2 to $-\infty$.

So the range is all real numbers less than 2.

(ii) $f(x) = x^2 + 2,\ x \in \mathbb{R}$

Since $x^2 \ge 0$, the smallest value of $f(x)$ is $0 + 2 = 2$. There is no upper bound.

So the range is $[2, \infty)$. …

The Core Idea: What Does "Range" Really Mean?

Before we dive into calculations, let's be clear about what we're actually looking for. The range of a function is the set of all possible output values — every $y$ that the function can actually produce when you feed it an allowed input $x$.

Think of it this way: the domain tells you which $x$ values you're allowed to plug in. The range tells you what $y$ values come out the other end. It's not about what the formula could theoretically give; it's about what it does give given the restrictions on $x$.

For each function, we'll ask: "If I put in every allowed $x$, what set of $y$ values do I get back?"

(i) $f(x) = 2 - 3x$, $x \in \mathbb{R}$, $x > 0$

1. Understand the restriction. The domain says $x$ can be any real number greater than 0. So $x$ can be 0.0001, 5, 1000, or any positive number — but never 0 itself and never a negative number.

2. See how the function behaves. This is a linear function with slope $-3$. As $x$ increases, $f(x)$ decreases. As $x$ gets closer to 0 from the positive side, $f(x)$ gets closer to $2 - 3(0) = 2$. But since $x$ can never actually be 0, $f(x)$ can never actually be 2 — it can only approach it from below.

3. Find the lower bound. As $x$ grows without bound ($x \to \infty$), $f(x) = 2 - 3x$ goes to $-\infty$. So there's no lower limit — the function can produce arbitrarily negative values.

4. Put it together. The function can produce any real number less than 2, but never 2 itself. In interval notation, that's $(-\infty, 2)$.

(ii) $f(x) = x^2 + 2$, $x$ is a real number

1. No restrictions here. The domain is all real numbers. So $x$ can be anything: negative, zero, positive, huge, tiny.

2. Think about the shape. $x^2$ is always non-negative. The smallest $x^2$ can be is 0 (when $x = 0$). So the smallest $f(x)$ can be is $0 + 2 = 2$. …