Mathematical-Reasoning · Unit 25 · Video 2 · Interactive Practice

Image and Preimage: Why Squaring Splits an Interval

IKey Formulas

Name	Definition	Membership test
Image of $A \subseteq S$	$f(A) = \{ f(x) \mid x \in A \} \subseteq T$	$y \in f(A)$ exactly when $y = f(x)$ for at least one $x \in A$
Preimage of $C \subseteq T$	$f^{-1}(C) = \{ x \in S \mid f(x) \in C \} \subseteq S$	$x \in f^{-1}(C)$ exactly when $f(x) \in C$
For $g(x) = x^2$	$g([1, 2]) = [1, 4]$	$g^{-1}([1, 4]) = [-2, -1] \cup [1, 2]$

Key Insight: The preimage $f^{-1}(C)$ is defined for every function. It uses $f$ alone, so no inverse function is needed: $g(x) = x^2$ has none, yet $g^{-1}([1, 4])$ is a well-defined set.

IIReading Images and Preimages off the Graph

An interval of inputs maps to one interval of heights, but a band of heights can pull back to two pieces.

Question a 1 b 2 c 1 d 4

IIIBuilding a Preimage One Input at a Time

A preimage needs no inverse: test each input, and keep it when its output lands in $C$ .

Function Target C x

IVQuiz Questions

Problem 1 · An Image That Crosses Zero

Given: $g: \mathbb{R} \to \mathbb{R}$ , $g(x) = x^2$ . Find the image $g([-1, 3])$ .

✅ Correct! The interval contains

0

, so the lowest height is

g(0) = 0

; the highest is

g(3) = 9

❌ Not quite. Squaring the end points misses the bottom of the parabola:

0 \in [-1, 3]

and

g(0) = 0

❌ Not quite. Every image under

g

consists of squares, and no square is negative.

❌ Not quite. That is the set of inputs whose squares lie in

[0, 9]

, a preimage. The image is a set of outputs.

Show solution

On $[-1, 0]$ the graph falls from $g(-1) = 1$ to $g(0) = 0$ ; on $[0, 3]$ it rises from $0$ to $g(3) = 9$ . Every height between $0$ and $9$ is reached, and no other height.

g([-1, 3]) = [0, 9].

Problem 2 · The Preimage of a Band

Given: $g(x) = x^2$ on $\mathbb{R}$ . Find the preimage $g^{-1}([4, 9])$ .

✅ Correct!

4 \le x^2 \le 9

holds for

2 \le x \le 3

and, symmetrically, for

-3 \le x \le -2

❌ Close, but… taking square roots forgets the negative inputs:

(-2.5)^2 = 6.25 \in [4, 9]

❌ Not quite. Inputs near

0

are too small:

g(1) = 1 \notin [4, 9]

, so the stretch

(-2, 2)

is left out.

❌ Not quite. That is the image

g([4, 9])

. The preimage collects inputs whose outputs lie in

[4, 9]

Show solution

We need every real $x$ with $4 \le x^2 \le 9$ . The condition $x^2 \le 9$ means $-3 \le x \le 3$ , and $x^2 \ge 4$ means $x \le -2$ or $x \ge 2$ . Both hold on two separate stretches:

g^{-1}([4, 9]) = [-3, -2] \cup [2, 3].

On the graph, these are the $x$ values where the parabola lies between the horizontal lines $y = 4$ and $y = 9$ .

Problem 3 · A Shifted Parabola

Given: $h: \mathbb{R} \to \mathbb{R}$ , $h(x) = x^2 - 2x$ , so $h(x) = (x - 1)^2 - 1$ . Find $h^{-1}([0, 3])$ .

✅ Correct!

0 \le (x-1)^2 - 1 \le 3

means

1 \le (x - 1)^2 \le 4

, so

x - 1 \in [-2, -1] \cup [1, 2]

❌ Close, but… that is only the right-hand piece.

h(-0.5) = 1.25 \in [0, 3]

, so inputs left of the vertex qualify too.

❌ Not quite. The lower bound matters here:

h(1) = -1 \notin [0, 3]

, so inputs near the vertex are left out.

❌ Not quite. A preimage is defined for every function; it never needs an inverse function. Test the inputs:

h(3) = 3 \in [0, 3]

Show solution

$x \in h^{-1}([0, 3])$ exactly when $0 \le (x-1)^2 - 1 \le 3$ , that is, $1 \le (x - 1)^2 \le 4$ .

Let $u = x - 1$ . Then $1 \le u^2 \le 4$ holds exactly for $u \in [-2, -1] \cup [1, 2]$ , as in the video's $g^{-1}([1, 4])$ . Adding $1$ gives

h^{-1}([0, 3]) = [-1, 0] \cup [2, 3].

Check: $h(-1) = 3$ , $h(0) = 0$ , $h(2) = 0$ and $h(3) = 3$ .

Problem 4 · What the Definitions Guarantee

Given: a function $f: S \to T$ , a subset $A \subseteq S$ and a subset $C \subseteq T$ . Which statement is true for every such $f$ , $A$ and $C$ ?

✅ Correct! The image collects outputs, so it lives in the codomain; the preimage collects inputs, so it lives in the domain.

❌ Not quite. The preimage uses only

f

g(x) = x^2

is not a bijection, yet

g^{-1}([1, 4]) = [-2, -1] \cup [1, 2]

❌ Not quite. A nonempty target can miss the range:

g^{-1}(\{-1\}) = \emptyset

for

g(x) = x^2

❌ Not quite.

y \in f(A)

says only that some element of

A

is sent to

y

. For

g(x) = x^2

4 \in g([1, 2])

, yet

g(-2) = 4

and

-2 \notin [1, 2]

Show solution

(c) is true straight from the definitions: $f(A) = \{ f(x) \mid x \in A \}$ consists of values of $f$ , which lie in $T$ , and $f^{-1}(C) = \{ x \in S \mid f(x) \in C \}$ is carved out of $S$ .

(a), (b) and (d) each fail for $g(x) = x^2$ : $g$ has no inverse function but has preimages; $g^{-1}(\{-1\}) = \emptyset$ ; and $-2$ is sent to $4 \in g([1, 2])$ without lying in $[1, 2]$ .

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