# Determining the Direct Image of A Set Examples

Given a set $E$ that is a subset of the domain $A$, recall the definition of the direct image of $E$ denoted $f(E)$:

Definition: Suppose $f : A \to B$ is a function where $A = D(f)$ and $R(f) \subseteq B$. If $E \subseteq A$, then we say the direct image of $E$ under the function $f$ is the subset of $B$ such that $f(E) = \{ f(x) : x \in E \}$. |

We will now look at some examples of determining the direct image of a subset of the domain of a function $f$ using the Extreme Value Theorem which says that a closed interval $I$ of a continuous function $f$ has an absolute maximum and absolute minimum on $I$.

## Example 1

**Let $f : \mathbb{R} \to \mathbb{R}$ be defined by the equation $f(x) = x + 1$. Let $E := \{ x \in \mathbb{R} : 1 ≤ x ≤ 2 \}$. Determine $f(E)$.**

From the extreme value theorem we acknowledge that on the interval $1 ≤ x ≤ 2$, the absolute minimum is $f(1) = 2$ and the absolute maximum is $f(2) = 3$. We thus get that the direct image of $E$ under $f$ is $f(E) = \{ y \in \mathbb{R} : 2 ≤ y ≤ 3 \}$.

## Example 2

**Let $f : \mathbb{R} \to \mathbb{R}$ be defined by the equation $f(x) = x^2$. Let $E = \{ x \in \mathbb{R} : -1 ≤ x ≤ 1 \}$. Determine $f(E)$.**

From the extreme value theorem we get that on the interval $-1 ≤ x ≤ 1$, the absolute minimum is $f(0) = 0$ and the absolute maximum is $f(-1) = f(1) = 1$. We thus get that the direct image of $E$ under $f$ is $f(E) = \{ y \in \mathbb{R} : 0 ≤ y ≤ 1 \}$.

## Example 3

**Let $f : \mathbb{R} \to \mathbb{R}$ be defined by the equation $f(x) = x^2 + 6x + 9$. Let $E = \{ x \in \mathbb{R} : x ≤ 1 \}$. Determine $f(E)$.**

We note that $f(x) = x^2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)^2$. We cannot fully use the extreme value theorem here since $x ≤ 1$ is not a closed interval. We note that $f(x) ≥ 0$, so $f$ is positive. A root exists when $x = -3$ and clearly $-3 ≤ 1$. This root is the minimum of the function, and $f(-3) = 0$. we thus get that $f(E) = \{ y \in \mathbb{R} : y ≥ 0 \}$.

## Example 4

**Let $g : \mathbb{R} \to \mathbb{R}$ and $f : \mathbb{R} \to \mathbb{R}$ be defined by $g(x) = x^2$ and $f(x) = x + 2$. Let $E = \{ x \in \mathbb{R} : 0 ≤ x ≤ 1 \}$. Determine $(g \circ f) ( E )$.**

We must exert some caution when dealing with compositions of functions. We note that $\mathbb{R} = R(f) \subseteq D(g) = \mathbb{R}$. So therefore $(g \circ f)(x) = g(f(x)) = g(x + 2) = (x + 2)^2$. So $(g \circ f) : \mathbb{R} \to \mathbb{R}$.

We note that by the extreme value theorem on the interval $0 ≤ x ≤ 1$, there is an absolute minimum at $(g \circ f)(0) = 4$ and an absolute maximum at $(g \circ f)(1) = 9$. Therefore $(g \circ f)(E) = \{ y \in \mathbb{R} : 4 ≤ y ≤ 9 \}$.