Let `f:R to R` be defined by
`f(x) = 5^(-|x|) - 5^(x) + sgn (e^(-x)) + 4` where sgn (x) denotes the signum function of x. Then

Options

A. f is one-to-one but not onto
B. f is one-to-one and onto
C. f is not onto but one-to-one
D. f is neither one-to-one nor onto

Answer

To analyze the function \( f: \mathbb{R} \to \mathbb{R} \) defined by

\[ f(x) = 5^{-|x|} - 5^x + \text{sgn}(e^{-x}) + 4, \]

we will determine whether the function is one-to-one (injective) and onto (surjective).

### Step 1: Understanding the Signum Function
The signum function \( \text{sgn}(x) \) is defined as:
- \( \text{sgn}(x) = 1 \) if \( x > 0 \)
- \( \text{sgn}(x) = 0 \) if \( x = 0 \)
- \( \text{sgn}(x) = -1 \) if \( x < 0 \)

Since \( e^{-x} > 0 \) for all \( x \in \mathbb{R} \), we have \( \text{sgn}(e^{-x}) = 1 \) for all \( x \). Therefore, we can simplify the function as:

\[ f(x) = 5^{-|x|} - 5^x + 1 + 4 = 5^{-|x|} - 5^x + 5. \]

### Step 2: Analyzing the Function for Positive and Negative \( x \)
We will analyze the function separately for positive and negative values of \( x \).

**Case 1: \( x \geq 0 \)**
For \( x \geq 0 \), \( |x| = x \), thus:

\[ f(x) = 5^{-x} - 5^x + 5. \]

**Case 2: \( x < 0 \)**
For \( x < 0 \), \( |x| = -x \), thus:

\[ f(x) = 5^{x} - 5^x + 5 = 5. \]

### Step 3: Determining Injectivity
To check if \( f \) is one-to-one, we need to analyze the behavior of \( f(x) \).

- For \( x < 0 \), \( f(x) = 5 \) is constant.
- For \( x \geq 0 \), we need to analyze \( f(x) = 5^{-x} - 5^x + 5 \).

To determine if \( f(x) \) is increasing or decreasing for \( x \geq 0 \), we compute the derivative:

\[ f'(x) = -\ln(5) \cdot 5^{-x} - \ln(5) \cdot 5^x. \]

Since both terms are negative for \( x \geq 0 \), we conclude that \( f'(x) < 0 \) for \( x \geq 0 \). Therefore, \( f(x) \) is decreasing in this interval.

Since \( f(x) \) is constant for \( x < 0 \) and decreasing for \( x \geq 0 \), it is not one-to-one.

### Step 4: Determining Surjectivity
To check if \( f \) is onto, we need to analyze the range of \( f(x) \).

- For \( x < 0 \), \( f(x) = 5 \).
- For \( x \geq 0 \), as \( x \to 0 \), \( f(0) = 5^{-0} - 5^0 + 5 = 5 \). As \( x \to \infty \), \( f(x) \to -\infty \) because \( -5^x \) dominates.

Thus, the range of \( f(x) \) for \( x \geq 0 \) is \( (-\infty, 5) \). Since \( f(x) = 5 \) for \( x < 0 \), the overall range of \( f \) is \( (-\infty, 5] \).

Since the codomain is \( \mathbb{R} \) and the range is \( (-\infty, 5] \), \( f \) is not onto.

### Conclusion
The function \( f(x) \) is neither one-to-one nor onto.