Answer
Yes. If \(f\) is onto and \(c \neq 0\), then \(c \cdot f\) is also onto.
Work Step by Step
**Problem Statement**
Let \(f: \mathbb{R} \to \mathbb{R}\) be a function, and let \(c\) be a nonzero real number. If \(f\) is onto (surjective), is the function \((c \cdot f)\), defined by \((c \cdot f)(x) = c \,f(x)\), also onto? Justify your answer.
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## Reasoning
1. **Definition of onto (surjective).**
A function \(f: \mathbb{R} \to \mathbb{R}\) is onto if, for **every** real number \(y\), there exists some \(x \in \mathbb{R}\) such that
\[
f(x) = y.
\]
2. **Show \((c \cdot f)\) is onto.**
Consider the function \((c \cdot f)\) defined by \((c \cdot f)(x) = c\,f(x)\).
- Let \(y\) be **any** real number. We want to find \(x \in \mathbb{R}\) such that \((c \cdot f)(x) = y\).
- This equation is
\[
c \, f(x) \;=\; y
\quad\Longrightarrow\quad
f(x) \;=\; \frac{y}{c}.
\]
- Because \(c \neq 0\), \(\frac{y}{c}\) is a well-defined real number.
- Since \(f\) is onto, there is an \(x \in \mathbb{R}\) for which \(f(x) = \tfrac{y}{c}\).
3. **Conclusion.**
Thus, for **every** real \(y\), we can solve \((c \cdot f)(x) = y\) by choosing \(x\) that satisfies \(f(x) = \tfrac{y}{c}\). Hence \((c \cdot f)\) is onto.
