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Is f(x) = eˣ Injective?

publish date: 2026/05/23 18:47:50.412002 UTC

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Consider \(f: (-\infty, \infty) \to (0, \infty)\) defined by \(f(x) = e^x\). Is this function one-one (injective)?

No — many values of x give the same output

Yes — if e^x₁ = e^x₂ then x₁ = x₂, so distinct inputs give distinct outputs

No — e^x can equal 0 for some x

Yes — only because the codomain is restricted to positive numbers

Correct Answer

Yes — if e^x₁ = e^x₂ then x₁ = x₂, so distinct inputs give distinct outputs

Explanation

\(e^x\) is strictly increasing: if \(e^{x_1} = e^{x_2}\) then \(e^{x_1-x_2} = 1 = e^0\), so \(x_1 - x_2 = 0\), giving \(x_1 = x_2\). Hence \(e^x\) is injective. With codomain \((0,\infty)\), it is also surjective, making it bijective from \((-\infty,\infty)\) onto \((0,\infty)\).

Reference

Introduction to Differential Calculus (Systematic Studies with Engineering Applications for Beginners) - 2012

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