📝 Chapter 1: Introduction to Functions

Assessment AS Learning — Practice Quiz

🔄 Not Graded — Unlimited Retakes

Purpose: Self-check your understanding. Use results to identify topics needing review before the unit test.

Score: 0 / 12

Topic 1.1–1.2 — Function Notation & Vertical Line Test

Question 1

Which relation is NOT a function?

\( y = 2x + 5 \)
\( y = x^2 - 4 \)
\( x = y^2 \)
\( y = \sqrt{x} \)

Solution:

\( x = y^2 \) fails the vertical line test — for \( x = 4 \), both \( y = 2 \) and \( y = -2 \) satisfy the equation. So one input maps to two outputs.

Question 2

If \( f(x) = 3x^2 - 2x + 1 \), find \( f(-2) \).

Solution:

\( f(-2) = 3(-2)^2 - 2(-2) + 1 = 3(4) + 4 + 1 = 12 + 4 + 1 = 17 \).

Topic 1.3 — Domain & Range

Question 3

State the domain of \( f(x) = \sqrt{x - 5} \).

\( \{x \in \mathbb{R}\} \)
\( \{x \in \mathbb{R} \mid x \geq 5\} \)
\( \{x \in \mathbb{R} \mid x > 5\} \)
\( \{x \in \mathbb{R} \mid x \leq 5\} \)

Solution:

The radicand must be non-negative: \( x - 5 \geq 0 \), so \( x \geq 5 \).

Question 4

For \( g(x) = \dfrac{1}{x - 3} \), the value of \( x \) excluded from the domain is:

Solution:

Denominator cannot be zero: \( x - 3 \neq 0 \), so \( x \neq 3 \).

Topic 1.4 — Inverse Functions

Question 5

If \( f(x) = 2x - 7 \), find \( f^{-1}(5) \).

Solution:

\( f^{-1}(x) = \dfrac{x + 7}{2} \), so \( f^{-1}(5) = \dfrac{5+7}{2} = 6 \). Equivalently, solve \( 2x - 7 = 5 \).

Question 6

The inverse of \( f(x) = x^2 \) is a function only if we restrict the domain to:

\( x \in \mathbb{R} \)
\( x \geq 0 \) (or \( x \leq 0 \))
\( x \neq 0 \)
\( x > 1 \)

Solution:

The inverse \( y = \pm\sqrt{x} \) is not a function. Restricting \( f \) to \( x \geq 0 \) (or \( x \leq 0 \)) makes the inverse a single-valued function.

Topic 1.5 — Transformations

Question 7

Describe the transformation in \( g(x) = -2 f(x - 3) + 4 \) compared to the parent \( y = f(x) \):

Vertical stretch by 2, reflection in x-axis, right 3, up 4
Horizontal stretch by 2, reflection in y-axis, left 3, down 4
Vertical compression by 2, right 3, down 4

Solution:

\( a = -2 \): vertical stretch by factor 2 and reflection in x-axis. \( d = 3 \): horizontal shift right 3. \( c = 4 \): vertical shift up 4.

Question 8

Apply the transformation \( y = (x + 2)^2 - 5 \) to the parent \( y = x^2 \). The vertex is at:

(2, 5)
(-2, -5)
(-2, 5)
(2, -5)

Solution:

Form is \( y = a(x - h)^2 + k \) with \( h = -2, k = -5 \). Vertex \( (-2, -5) \).

Topic 1.6 — Even & Odd Functions

Question 9

Is \( f(x) = x^4 - 3x^2 + 1 \) even, odd, or neither?

Even
Odd
Neither

Solution:

\( f(-x) = (-x)^4 - 3(-x)^2 + 1 = x^4 - 3x^2 + 1 = f(x) \). Even function (y-axis symmetric).

Question 10

Is \( g(x) = x^3 - 4x \) even, odd, or neither?

Even
Odd
Neither

Solution:

\( g(-x) = -x^3 + 4x = -(x^3 - 4x) = -g(x) \). Odd function (origin symmetric).

Mixed Application

Question 11

Given \( f(x) = x^2 - 4 \) and \( g(x) = 2x + 1 \), evaluate \( f(g(2)) \).

Solution:

\( g(2) = 2(2) + 1 = 5 \). Then \( f(5) = 5^2 - 4 = 21 \).

Question 12

For \( f(x) = 2x + 5 \), find \( \dfrac{f(a + h) - f(a)}{h} \).

2
\( 2a + 5 \)
\( 2h \)
\( h + 5 \)

Solution:

\( f(a+h) - f(a) = [2(a+h) + 5] - [2a + 5] = 2h \). Divide by \( h \): result is 2 (the slope of the line).