Back to Mathematics syllabus

Range of a Function

CSCA Range of a Function study guide organized around the publicly available CSCA syllabus. Practice Mathematics questions on aicsca.com.

Before planning this topic, check the CSCA Exam Guide 2026 for exam dates, registration, fees, and subject requirements.

Syllabus Alignment

This study guide is organized around the publicly available CSCA syllabus for international undergraduate applicants.

Who It Is For

International students preparing for CSCA Math, Physics, Chemistry, or Chinese exams.

Related Practice

CSCA Practice · Go to questions

Past papers and worked video solutions · Timed mock exams · All subject video lessons

Practice by Topic

Jump from this tutorial to filtered practice questions for the same knowledge point.

Tutorial Content

Range of a Function

1. Concept

The **range** of a function $y=f(x)$ is the set of all possible output values $y$ corresponding to the input values $x$ in the domain $D$. It is denoted as $R_f = \{ f(x) | x \in D \}$.

**Intuitive Understanding**:

If you project the graph of a function horizontally onto the y-axis, the interval covered on the y-axis represents the range.

---

2. Four Common Methods for Finding Range

There is no single formula for finding the range; the method depends on the function's form.

#### (1) Completing the Square (For Quadratics)

**Target**: Quadratic functions or composite functions involving quadratics.

**Method**: Convert $ax^2+bx+c$ into the vertex form $a(x-h)^2+k$. Determine the range based on the sign of $a$ (opening direction) and the vertex.

#### (2) Separation of Constants / Inverse Method

**Target**: Linear fractional functions $y = \frac{ax+b}{cx+d}$.

**Method**:

* **Separation**: Rewrite the numerator to match the denominator, e.g., $y = \frac{2x+3}{x-1} = 2 + \frac{5}{x-1}$. Since $\frac{5}{x-1} \neq 0$, $y \neq 2$.

* **Inverse Method**: Solve for $x$ in terms of $y$. The restriction on $y$ (usually denominator $\neq 0$) gives the range.

#### (3) Discriminant Method

**Target**: Rational functions with quadratic terms, like $y = \frac{x}{x^2+1}$.

**Principle**: Rearrange the function into a quadratic equation regarding $x$: $A(y)x^2 + B(y)x + C(y) = 0$. Since $x$ must be real, the discriminant $\Delta \ge 0$. Solve the resulting inequality for $y$.

#### (4) Monotonicity

**Target**: Functions that are strictly increasing or decreasing on a given interval.

**Method**: Evaluate the function at the endpoints of the interval. If $f(x)$ increases on $[a, b]$, the range is $[f(a), f(b)]$.

---

3. Classic Examples

**Example 1: Completing the Square**

Find the range of $f(x)=x^2-4x+5$ on $\mathbb{R}$.

**Solution**:

$f(x) = (x-2)^2 + 1$.

Since $(x-2)^2 \ge 0$, $f(x) \ge 1$.

**Range**: $[1, +\infty)$.

**Example 2: Discriminant Method**

Find the range of $y = \frac{x}{x^2+1}$.

**Solution**:

Rearrange to $yx^2 - x + y = 0$.

1. If $y=0$, then $x=0$ (Valid).

2. If $y \neq 0$, treat as quadratic in $x$. Since $x \in \mathbb{R}$, $\Delta \ge 0$.

$\Delta = 1 - 4y^2 \ge 0 \Rightarrow y^2 \le \frac{1}{4} \Rightarrow -\frac{1}{2} \le y \le \frac{1}{2}$.

**Range**: $[-\frac{1}{2}, \frac{1}{2}]$.

**Example 3: Monotonicity**

Find the range of $f(x)=\sqrt{x-1} + 2$ on $[1, 5]$.

**Solution**:

The function is increasing on $[1, 5]$.

Min $f(1) = 2$, Max $f(5) = 4$.

**Range**: $[2, 4]$.

---

4. Common Pitfalls

* **Ignoring the Domain**: E.g., for $y=x^2-4x+5$ on $x \in [0, 1]$, the minimum is NOT at the vertex ($x=2$) because 2 is outside the domain. You must check the endpoints.

* **Discriminant Method Oversight**: Always check the case where the leading coefficient is 0 (i.e., when $y=0$ in the quadratic equation step).