# Image of Mapping/Examples/Image of x^2-4x+5

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## Example of Image of Element under Mapping

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

- $\forall x \in \R: \map f x = x^2 - 4 x + 5$

The image of $f$ is the unbounded closed interval:

- $\Img f = \hointr 1 \to$

and so $f$ is not a surjection.

### Graphical Representation of $\map f x = x^2 - 4 x + 5$

Image of Mapping/Examples/Image of x^2-4x+5/Graph

## Proof

By differentiating $x^2 - 4 x + 5$ twice with respect to $x$:

- $f' = 2 x - 4$

- $f' = 2 x - 4$

\(\ds f'\) | \(=\) | \(\ds 2 x - 4\) | ||||||||||||

\(\ds f''\) | \(=\) | \(\ds 2\) |

Equating $f'$ to $0$, a stationary point is found at $x = 2$.

Inspecting the sign of $f''$, it is noted that $f'$ is increasing everywhere.

Hence the stationary point at $x = 2$ is a minimum of $\Img f$.

This is the only stationary point, so it can be stated that **the** minimum of $f$ occurs at $x = 2$.

We have that:

- $f \paren 2 = 2^2 - 4 \times 2 + 5 = 4 - 8 + 5 = 1$

As $f$ is strictly increasing on $x > 2$ and strictly decreasing on $x < 2$, it is seen that $f$ is unbounded above.

Thus:

- $\Img f = \hointr 1 \to$

$\blacksquare$

## Also see

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $4$: Mappings: Exercise $12 \ \text{(ii)}$