If `f(x)={{:(,x^(2)+1,0 le x lt 1),(,-3x+5, 1 le x le 2):}`

Options

A. It has a relative minimum at x = 1
B. It has a relative maximum at x =1
C. It is not continuous at x = 1
D. None of these

Answer

To solve the given piecewise function \( f(x) \), we need to analyze the function in two segments based on the defined intervals.

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} x^2 + 1 & \text{for } 0 \leq x < 1 \\ -3x + 5 & \text{for } 1 \leq x \leq 2 \end{cases} \]

### Step 2: Analyze the first segment \( f(x) = x^2 + 1 \)
For \( 0 \leq x < 1 \):
- This is a quadratic function which opens upwards (since the coefficient of \( x^2 \) is positive).
- The minimum value occurs at \( x = 0 \):
\[ f(0) = 0^2 + 1 = 1 \]
- At \( x = 1 \) (not included in this segment):
\[ f(1) = 1^2 + 1 = 2 \]

### Step 3: Analyze the second segment \( f(x) = -3x + 5 \)
For \( 1 \leq x \leq 2 \):
- This is a linear function with a negative slope.
- At \( x = 1 \):
\[ f(1) = -3(1) + 5 = 2 \]
- At \( x = 2 \):
\[ f(2) = -3(2) + 5 = -1 \]

### Step 4: Determine continuity at \( x = 1 \)
To check if the function is continuous at \( x = 1 \):
- The left-hand limit as \( x \) approaches 1 from the left:
\[ \lim_{x \to 1^-} f(x) = f(1) = 2 \]
- The right-hand limit as \( x \) approaches 1 from the right:
\[ \lim_{x \to 1^+} f(x) = f(1) = 2 \]
- Since both limits are equal and \( f(1) = 2 \), the function is continuous at \( x = 1 \).

### Step 5: Identify maximum and minimum values
- The minimum value in the first segment occurs at \( x = 0 \):
\[ f(0) = 1 \]
- The maximum value occurs at \( x = 1 \):
\[ f(1) = 2 \]
- The value at \( x = 2 \) is:
\[ f(2) = -1 \]

### Conclusion
- The maximum value of \( f(x) \) occurs at \( x = 1 \) with \( f(1) = 2 \).
- The minimum value occurs at \( x = 2 \) with \( f(2) = -1 \).

### Final Answer
The maximum value of the function is at \( x = 1 \) where \( f(1) = 2 \).