The function `f :R to R` defined by `f (x) = e ^(x)` is

Options

A. Onto
B. Many-one
C. One-one and into
D. Many one and onto

Answer

To determine the properties of the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = e^x \), we need to analyze whether it is one-to-one (injective) and onto (surjective).

### Step 1: Check if the function is one-to-one (1-1)

A function is one-to-one if for every pair of distinct inputs, the outputs are also distinct. This can be checked using the horizontal line test or by showing that if \( f(a) = f(b) \), then \( a = b \).

1. Assume \( f(a) = f(b) \).
2. This means \( e^a = e^b \).
3. Taking the natural logarithm on both sides, we get \( \ln(e^a) = \ln(e^b) \).
4. This simplifies to \( a = b \).

Since we have shown that \( f(a) = f(b) \) implies \( a = b \), the function \( f(x) = e^x \) is one-to-one.

### Step 2: Check if the function is onto (onto)

A function is onto if every element in the codomain has a pre-image in the domain. In this case, we need to check if every real number \( y \) has a corresponding \( x \) such that \( f(x) = y \).

1. The function \( f(x) = e^x \) produces outputs that are always positive, i.e., \( f(x) > 0 \) for all \( x \in \mathbb{R} \).
2. The range of \( f(x) = e^x \) is \( (0, \infty) \).
3. The codomain is \( \mathbb{R} \), which includes negative numbers and zero.

Since there are no \( x \) values such that \( f(x) = y \) for \( y \leq 0 \), the function is not onto.

### Conclusion

The function \( f(x) = e^x \) is one-to-one but not onto. Therefore, we can conclude:

- The function is **1-1** (injective).
- The function is **not onto** (not surjective).

### Final Answer

The function \( f(x) = e^x \) is one-to-one (1-1) and not onto (not surjective).

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