Exercise · ZW-IMAGE-AND-PREIMAGE-009 · ★★★ Advanced

Let \(f\colon\mathbb{R}\to\mathbb{R}\), \(f(x)=x^2-2x\). What is \(f^{-1}(\{-1\})\)?

Choose one option.

Show answer
  • C. \(\{1\}\)

Explanation

Solve \(x^2-2x=-1\), which is equivalent to \((x-1)^2=0\), so \(x=1\) is the only solution and the preimage is \(\{1\}\). The other options fail on substitution: for example \(f(-1)=3\), not \(-1\).

Study Set theory in ZWMath →

Review the concept: Image and preimage

Given \(f\colon A\to B\), the image of \(X\subseteq A\) is the set of values \(f(x)\) with \(x\in X\), and the preimage of \(Y\subseteq B\) is the set of \(a\in A\) whose value lies in \(Y\).

\[f(X)=\{f(x)\colon x\in X\},\qquad f^{-1}(Y)=\{a\in A\colon f(a)\in Y\}\]

More practice problems: image and preimage

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