# Image of Subset under Mapping/Examples/Image of -3 to 2 under x^4-1

Jump to navigation
Jump to search

## Example of Image of Element under Mapping

Let $f: \R \to \R$ be the mapping defined as:

- $\forall x \in \R: f \paren x = x^4 - 1$

The image of the closed interval $\closedint {-3} 2$ is:

- $f \sqbrk {\paren {\closedint {-3} 2} } = \closedint {-1} {80}$

## Proof

Trivially, by differentiating $x^4 - 1$ with respect to $x$:

- $f' = 4 x^3$

and equating $f'$ to $0$, the minimum of $\Img f$ is seen to occur at $f \paren 0 = -1$.

As $0 \in \closedint {-3} 2$, it can be seen that the minimum of $f \sqbrk {\paren {\closedint {-3} 2} }$ is $-1$.

As $f$ is strictly increasing on $x > 0$ and strictly decreasing on $x < 0$, it suffices to inspect the images of the endpoints $-3$ and $2$.

Thus:

- $f \paren {-3} = \paren {-3}^4 - 1 = 81 - 1 = 80$

- $f \paren 2 = 2^4 - 1 = 16 - 1 = 15$

The result follows.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $4$: Mappings: Exercise $3 \ \text {(ii)}$