If `f(x)=e^(1-x) `then f(x) is

Options

A. increasing in `[-1//2,1]`
B. decreasing in R
C. increasing in R
D. decreasing in `[-1//2,1]`

Answer

To determine the monotonicity of the function \( f(x) = e^{1-x} \), we will follow these steps:

### Step 1: Differentiate the function
First, we need to find the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(e^{1-x}) \]

Using the chain rule, we differentiate \( e^{u} \) where \( u = 1 - x \):

\[ f'(x) = e^{1-x} \cdot \frac{d}{dx}(1-x) = e^{1-x} \cdot (-1) = -e^{1-x} \]

### Step 2: Analyze the sign of the derivative
Next, we analyze the sign of the derivative \( f'(x) \):

\[ f'(x) = -e^{1-x} \]

Since the exponential function \( e^{1-x} \) is always positive for all real numbers \( x \), we have:

\[ -e^{1-x} < 0 \]

This means that \( f'(x) < 0 \) for all \( x \in \mathbb{R} \).

### Step 3: Conclusion about monotonicity
Since the derivative \( f'(x) \) is negative for all \( x \), it indicates that the function \( f(x) \) is decreasing on the entire real line.

Thus, we conclude that:

\[ f(x) \text{ is decreasing on } (-\infty, \infty). \]

### Final Answer
The function \( f(x) = e^{1-x} \) is decreasing for all \( x \in \mathbb{R} \).

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