If `f(x)={{:(x",",0lexle1),(2-e^(x-1)",",1ltxle2),(x-e",",2ltxle3):}` and `g'(x)=f(x), x in [1,3]`, then```

Options

A. g(x) has no local maxima
B. g(x) has no local minima
C. g(x) has local maxima at `x=1+ln2` and local minima at x = e
D. g(x) has local minima at `x=1+ln2` and local maxima at x = e

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined piecewise and then find the critical points of the function \( g(x) \) whose derivative is given by \( g'(x) = f(x) \) for \( x \in [1, 3] \).

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} 0 & \text{if } 0 \leq x \leq 1 \\ 2 - e^{(x-1)} & \text{if } 1 < x \leq 2 \\ x - e & \text{if } 2 < x \leq 3 \end{cases} \]

### Step 2: Find the derivative \( g'(x) = f(x) \)
We know that \( g'(x) = f(x) \). We will analyze \( f(x) \) in the intervals \( (1, 2) \) and \( (2, 3) \).

### Step 3: Analyze \( f(x) \) in the interval \( (1, 2) \)
In the interval \( (1, 2) \):
\[ f(x) = 2 - e^{(x-1)} \]
To find the critical points, we need to find where \( f(x) = 0 \):
\[ 2 - e^{(x-1)} = 0 \implies e^{(x-1)} = 2 \implies x - 1 = \ln(2) \implies x = 1 + \ln(2) \]

### Step 4: Analyze \( f(x) \) in the interval \( (2, 3) \)
In the interval \( (2, 3) \):
\[ f(x) = x - e \]
To find the critical points, we set \( f(x) = 0 \):
\[ x - e = 0 \implies x = e \]

### Step 5: Determine the nature of critical points
Now we need to check the nature of the critical points \( x = 1 + \ln(2) \) and \( x = e \).

1. **At \( x = 1 + \ln(2) \)**:
- For \( x < 1 + \ln(2) \), \( f(x) > 0 \) (since \( 2 - e^{(x-1)} > 0 \)).
- For \( x > 1 + \ln(2) \), \( f(x) < 0 \) (since \( 2 - e^{(x-1)} < 0 \)).
- Therefore, \( x = 1 + \ln(2) \) is a point of maxima.

2. **At \( x = e \)**:
- For \( x < e \), \( f(x) < 0 \) (since \( x - e < 0 \)).
- For \( x > e \), \( f(x) > 0 \) (since \( x - e > 0 \)).
- Therefore, \( x = e \) is a point of minima.

### Conclusion
The critical points are:
- Maximum at \( x = 1 + \ln(2) \)
- Minimum at \( x = e \)