If $f:R\rightarrow R$, where R is set of real numbers, such that $f(x)=e^{x}$, then f is:

Question

Accepted Answer

To determine why Option 2, "injective only," is the correct answer for the function $ f: \mathbb{R} \rightarrow \mathbb{R} $ defined by $ f(x) = e^x $, we need to analyze the properties of the function in terms of injectivity and surjectivity.

1. **Injectivity**: A function is injective (or one-to-one) if different inputs map to different outputs. For the function $ f(x) = e^x $, consider two arbitrary real numbers $ a $ and $ b $. If $ f(a) = f(b) $, then $ e^a = e^b $. The exponential function $ e^x $ is strictly increasing, meaning that if $ e^a = e^b $, it must be the case that $ a = b $. Therefore, $ f(x) = e^x $ is injective because it maps distinct inputs to distinct outputs.

2. **Surjectivity**: A function is surjective (or onto) if every element in the codomain (in this case, the set of all real numbers $\mathbb{R}$) is the image of at least one element from the domain. For $ f(x) = e^x $, the range is $ (0, \infty) $, which means it only takes positive real values. Since there are real numbers (specifically, non-positive numbers) that are not in the range of $ f(x) = e^x $, the function is not surjective when considering the codomain as all real numbers $\mathbb{R}$.

3. **Bijectivity**: A function is bijective if it is both injective and surjective. Since $ f(x) = e^x $ is not surjective over $\mathbb{R}$, it cannot be bijective.

4. **Neither Surjective Nor Injective**: This option would imply that the function is neither one-to-one nor onto. However, as established, $ f(x) = e^x $ is injective, so this option is incorrect.

In conclusion, the function $ f(x) = e^x $ is injective because it maps distinct inputs to distinct outputs, but it is not surjective over the set of all real numbers because it does not cover the entire real line. Therefore, Option 2, "injective only," is the correct answer.