If `f(x) = {{:(e ^(x),,"," 0 le x lt 1 ,, ""), (2- e^(x - 1),,"," 1 lt x le 2,, and g(x) = int_(0)^(x) f(t ) dt","),( x- e,,"," 2lt x le 3 ,, ""):}`
` x in [ 1, 3 ] `, then

Options

A. g(x) has local maxima at `x=1+log_(e)2` and local minima at x=e
B. f(x) has local maxima at x=1 and local minima at x=2
C. g(x) has no local minima
D. f(x) has no local maxima

Answer

To solve the problem, we need to analyze the piecewise function \( f(x) \) and its integral \( g(x) \) over the interval \( [1, 3] \). We will find the local maxima and minima of \( g(x) \).

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is defined as follows:
- For \( 0 \leq x < 1 \): \( f(x) = e^x \)
- For \( 1 < x \leq 2 \): \( f(x) = 2 - e^{x-1} \)
- For \( 2 < x \leq 3 \): \( f(x) = x - e \)

### Step 2: Define the integral \( g(x) \)
The function \( g(x) \) is defined as:
\[ g(x) = \int_0^x f(t) \, dt \]
We need to evaluate \( g(x) \) for \( x \) in the interval \( [1, 3] \).

### Step 3: Find \( g(x) \) for \( x \) in different intervals
1. **For \( 1 < x \leq 2 \)**:
\[ g(x) = \int_0^1 f(t) \, dt + \int_1^x f(t) \, dt \]
- Calculate \( \int_0^1 e^t \, dt = [e^t]_0^1 = e - 1 \)
- Calculate \( \int_1^x (2 - e^{t-1}) \, dt \):
\[ = \int_1^x 2 \, dt - \int_1^x e^{t-1} \, dt = 2(x - 1) - [e^{t-1}]_1^x = 2(x - 1) - (e^{x-1} - 1) \]
- Thus,
\[ g(x) = (e - 1) + 2(x - 1) - (e^{x-1} - 1) = e - 1 + 2x - 2 - e^{x-1} + 1 = e + 2x - 2 - e^{x-1} \]

2. **For \( 2 < x \leq 3 \)**:
\[ g(x) = \int_0^1 e^t \, dt + \int_1^2 (2 - e^{t-1}) \, dt + \int_2^x (t - e) \, dt \]
- We already calculated the first two integrals:
\[ g(2) = e - 1 + (2 - e) = e + 1 \]
- Now calculate \( \int_2^x (t - e) \, dt \):
\[ = \left[\frac{t^2}{2} - et\right]_2^x = \left(\frac{x^2}{2} - ex\right) - \left(\frac{4}{2} - 2e\right) = \frac{x^2}{2} - ex - 2 + 2e \]
- Thus,
\[ g(x) = (e + 1) + \left(\frac{x^2}{2} - ex - 2 + 2e\right) = \frac{x^2}{2} - ex + e - 1 \]

### Step 4: Find \( g'(x) \)
To find local maxima and minima, we differentiate \( g(x) \):
1. For \( 1 < x \leq 2 \):
\[ g'(x) = 2 - e^{x-1} \]
2. For \( 2 < x \leq 3 \):
\[ g'(x) = x - e \]

### Step 5: Set \( g'(x) = 0 \) to find critical points
1. **For \( 1 < x \leq 2 \)**:
\[ 2 - e^{x-1} = 0 \implies e^{x-1} = 2 \implies x - 1 = \ln(2) \implies x = 1 + \ln(2) \]

2. **For \( 2 < x \leq 3 \)**:
\[ x - e = 0 \implies x = e \]

### Step 6: Determine the nature of critical points
- For \( x = 1 + \ln(2) \):
- Test values around \( 1 + \ln(2) \) to see if \( g'(x) \) changes from positive to negative (local maximum).
- For \( x = e \):
- Test values around \( e \) to see if \( g'(x) \) changes from negative to positive (local minimum).

### Conclusion
- Local maximum at \( x = 1 + \ln(2) \)
- Local minimum at \( x = e \)