Preimage of Subset under Mapping/Examples/Preimage of -2 to 0 under x^2-x-2
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Example of Preimage 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 preimage of the closed interval $\closedint {-2} 0$ is:
- $f^{-1} \closedint {-2} 0 = \closedint {-1} 0 \cup \closedint 1 2$
Proof
Trivially, by differentiating $x^2 - x - 2$ with respect to $x$:
- $f' = 2 x - 1$
and equating $f'$ to $0$, the minimum of $\Img f$ is seen to occur at $\map f {\dfrac 1 2} = -\dfrac 9 4$.
We see that $\closedint {-2} 0$ is well within the codomain of $f$.
Hence we should be able to solve for $\map f x = -2$ and $\map f x = 0$ and get two values for each.
So:
\(\ds \map f x\) | \(=\) | \(\ds -2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 - x - 2\) | \(=\) | \(\ds -2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 - x\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \paren {x - 1}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 0 \text { or } x = 1\) |
Then:
\(\ds \map f x\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 - x - 2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x - 2} \paren {x + 1}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 2 \text { or } x = -1\) |
The result follows.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions