The equation `e^x-x-1=0` has apart from x=0

Options

A. one real root
B. two real roots
C. no real root
D. infinite number of real roots

Explanation

To solve the equation \( e^x - x - 1 = 0 \) and determine if there are any roots apart from \( x = 0 \), we can follow these steps:

### Step 1: Define the function
We start by defining a function based on the given equation:
\[ f(x) = e^x - x - 1 \]

### Step 2: Evaluate the function at \( x = 0 \)
Next, we evaluate the function at \( x = 0 \):
\[ f(0) = e^0 - 0 - 1 = 1 - 0 - 1 = 0 \]
This shows that \( x = 0 \) is indeed a root of the equation.

### Step 3: Analyze the behavior of the function
To check for other roots, we need to analyze the behavior of the function \( f(x) \). We can do this by finding the derivative:
\[ f'(x) = e^x - 1 \]

### Step 4: Determine critical points
Set the derivative equal to zero to find critical points:
\[ f'(x) = 0 \implies e^x - 1 = 0 \implies e^x = 1 \implies x = 0 \]
This indicates that \( x = 0 \) is the only critical point.

### Step 5: Analyze the sign of the derivative
Now, we check the sign of \( f'(x) \):
- For \( x < 0 \), \( e^x < 1 \) so \( f'(x) < 0 \) (the function is decreasing).
- For \( x > 0 \), \( e^x > 1 \) so \( f'(x) > 0 \) (the function is increasing).

### Step 6: Conclusion about the roots
Since \( f(x) \) is decreasing for \( x < 0 \) and increasing for \( x > 0 \), and we found that \( f(0) = 0 \), it indicates that \( x = 0 \) is the only root. Therefore, there are no other real roots apart from \( x = 0 \).

### Final Answer
The equation \( e^x - x - 1 = 0 \) has only one root, which is \( x = 0 \), and no other real roots.

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