Question

Prove that the function \( f : \mathbb{R} \to \mathbb{R}^+ \) defined by \( f(x) = e^x \) is one-one.

Hint:

To prove a function is one-one, assume \( f(x_1) = f(x_2) \) and show that this implies \( x_1 = x_2 \).

Answer 1

A function \( f(x) \) is said to be one-one (injective) if for every \( f(x_1) = f(x_2) \), we have \( x_1 = x_2 \). Let us assume \( f(x_1) = f(x_2) \), i.e., \[ e^{x_1} = e^{x_2} \] Taking the natural logarithm of both sides: \[ \ln(e^{x_1}) = \ln(e^{x_2}) \] Since \( \ln(e^x) = x \), we get: \[ x_1 = x_2 \] Thus, we have proved that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \). Therefore, the function \( f(x) = e^x \) is one-one. Final Answer: The function \( f(x) = e^x \) is one-one.