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

Example of Image of Subset under Mapping
Let $f: \R \to \R$ be the mapping defined as:


 * $\forall x \in \R: \map f x = x^2 - x - 2$

The image of the open interval $\openint {-1} 1$ is:


 * $f \sqbrk {\openint {-1} 1} = \hointr {-\dfrac 9 4} 0$

Proof
Trivially, by differentiating $x^2 - x - 2$ $x$:
 * $f' = 2 x - 1$

and equating $f'$ to $0$, the minimum of $\Img f$ is seen to occur at $\map f {\dfrac 1 2} = -\dfrac 9 4$.

As $\dfrac 1 2 \in \closedint {-1} 1$, it can be seen that the minimum of $f \sqbrk {\openint {-1} 1}$ is $-\dfrac 9 4$.

As $f$ is strictly increasing on $x > 0$ and strictly decreasing on $x < 0$, it suffices to inspect the images of the endpoints $-1$ and $1$.

Thus:
 * $\map f {-1} = \paren {-1}^2 - \paren {-1} - 2 = 0$


 * $\map f 1 = 1^2 - 1 - 2 = -2$

The result follows.