Determining the Inverse Image of A Set Examples

# Determining the Inverse Image of A Set Examples

Recall, given a set $H$ that is a subset of the codomain $B$, the inverse image of the set $H$ denoted $f^{-1} (H)$ is defined as follows:

 Definition: If $f : A \to B$ is a function where $H \subseteq B$ then the inverse image of $H$ under the function $f$ is defined to be the set $f^{-1}(H) = \{ x \in A : f(x) \in H \}$ containing elements $x \in A$ such that $f(x) \in H$.

We will now look at some examples of determining the inverse image of a set.

## Example 1

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by the equation $f(x) = 3x$. If $H \subseteq B = \mathbb{R}$ and $H := \{ y \in \mathbb{R} : 3 ≤ y ≤ 6 \}$, then determine the inverse image $f^{-1} (H)$.

To find the inverse image of the set $H$, $f^{-1} (H)$, we need to find $x \in A$ such that $f(x) \in H$. We note that if $x = 1$ then $f(1) = 3$ and similarly if $x = 2$ then $f(2) = 6$. So on the interval $1 ≤ x ≤ 2$ we have that $3 ≤ f(x) = 3x ≤ 6$. This is the only interval where this is true, so $f^{-1} (H) = \{ x \in \mathbb{R} : 1 ≤ x ≤ 2$.

We should note that if we take the inequality $3 ≤ 3x ≤ 6$ and divide each part by $3$ we obtain an interval to which $x$ is restricted for our inverse image.

## Example 2

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by the equation $f(x) = x^2 + 1$. Let $H \subseteq B = \mathbb{R}$ and $H := \{ y \in \mathbb{R} : 5 ≤ y ≤ 10 \}$, then determine the inverse image $f^{-1} (H)$.

We want to find values of $x$ such that $5 ≤ x^2 + 1 ≤ 10$. Subtracting 1 from each part of this inequality we get that $4 ≤ x^2 ≤ 9$

But this only happens if $-3 ≤ x ≤ -2$ or $2 ≤ x ≤ 3$. So we get that $f^{-1} (H) = \{ x \in \mathbb{R} : -3 ≤ x ≤ -2 \: \mathrm{or} \: 2 ≤ x ≤ 3 \}$.

## Example 3

Let $p : \mathbb{R} \to \mathbb{R}$ be defined by the equation $p(x) = x^2 + x$. Let $H \subseteq B = \mathbb{R}$ and $H := \{ y \in \mathbb{R} : 6 ≤ y ≤ 12 \}$, the determine the inverse image $p^{-1} (H)$.

We need to find values of $x$ such that $6 ≤ x^2 + x ≤ 12$. We note that $p(2) = 6$ and $p(3) = 12$. On the interval $2 ≤ x ≤ 3$, the absolute minimum is $p(2) = 6$ and the absolute maximum is $p(3) = 12$. So we have found one interval in the inverse image $p^{-1} (H)$.

Now since $p$ is a parabola we suspect that there are two sets of intervals such that $6 ≤ y ≤ 12$. We note that $6 = p(x) = x^2 + x$ if $x = 2$ (as we have verified) or $x = -3$. Similarly we note that $12 = p(x) = x^2 + x$ if $x = 3$ (as we have verified) or $x = -4$.

Therefore on the interval $-4 ≤ x ≤ -3$, $6 ≤ x^2 + x ≤ 12$ and so then $f^{-1} (H) = \{ x \in \mathbb{R} : -4 ≤ x ≤ -3 \: \mathrm{or} \: 2 ≤ x ≤ 3 \}$.