Answer
Yes. If \(f\) is one-to-one and \(c \neq 0\), then \(c \cdot f\) is also one-to-one.
Work Step by Step
**Problem Statement**
Let \(f: \mathbb{R} \to \mathbb{R}\) be a function and let \(c\) be a nonzero real number. Suppose \(f\) is one-to-one (injective). Is the function \((c \cdot f)\), defined by \((c \cdot f)(x) = c \,f(x)\), also one-to-one? Justify your answer.
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## Reasoning
1. **Definition of one-to-one (injective).**
A function \(f\) is one-to-one if, for every \(x_1, x_2\in\mathbb{R}\),
\[
f(x_1) = f(x_2)
\quad\Longrightarrow\quad
x_1 = x_2.
\]
2. **Show \((c \cdot f)\) inherits injectivity.**
Consider the function \((c \cdot f)\) defined by \((c \cdot f)(x) = c\,f(x)\). Assume \((c \cdot f)(x_1) = (c \cdot f)(x_2)\). Then
\[
c\,f(x_1) \;=\; c\,f(x_2).
\]
Since \(c \neq 0\), we can divide both sides by \(c\), giving
\[
f(x_1) \;=\; f(x_2).
\]
Because \(f\) is one-to-one, \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). Therefore, whenever \((c \cdot f)(x_1) = (c \cdot f)(x_2)\), it must follow that \(x_1 = x_2\).
3. **Conclusion.**
Hence \((c \cdot f)\) is one-to-one.
