The function ` f(x) = 7 + x - e^(x)` is strictly increasing in the interval

Options

A. `(1, infty)`
B. `(infty, 1)`
C. `(0, infty)`
D. `(- infty, 0)`

Answer

To determine the interval in which the function \( f(x) = 7 + x - e^x \) is strictly increasing, we will follow these steps:

### Step 1: Find the derivative of the function
To check if the function is strictly increasing, we first need to find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(7 + x - e^x) \]
The derivative of \( 7 \) is \( 0 \), the derivative of \( x \) is \( 1 \), and the derivative of \( e^x \) is \( e^x \). Therefore, we have:

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

### Step 2: Set the derivative greater than zero
For the function to be strictly increasing, its derivative must be greater than zero:

\[ f'(x) > 0 \]
This leads to the inequality:

\[ 1 - e^x > 0 \]

### Step 3: Solve the inequality
Rearranging the inequality gives:

\[ 1 > e^x \]
or equivalently,

\[ e^x < 1 \]

### Step 4: Analyze the exponential function
The exponential function \( e^x \) is less than \( 1 \) when \( x < 0 \). This is because the graph of \( e^x \) intersects the line \( y = 1 \) at \( x = 0 \) and is below \( 1 \) for all negative values of \( x \).

### Step 5: Conclusion
Thus, the function \( f(x) \) is strictly increasing for:

\[ x < 0 \]

### Final Answer
The function \( f(x) = 7 + x - e^x \) is strictly increasing in the interval \( (-\infty, 0) \).

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