Analyzing the Given Piecewise Function

The problem asks us to consider a piecewise function defined as:

\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {3{{\rm{x}}^2} + 12{\rm{x}} - 1,{\rm{\;\;}} - 1 \le {\rm{x}} \le 2}\\ {37 - {\rm{x}},{\rm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}2 < {\rm{x}} \le 3} \end{array}} \right.\)

We need to evaluate the correctness of three statements regarding this function at the point x = 2: continuity, greatest value, and differentiability.

Statement 1: f(x) is continuous at x = 2

For a function to be continuous at a point \(x=c\), three conditions must be met:

1. \(f(c)\) must be defined.
2. The limit as \(x\) approaches \(c\) must exist (\(\lim_{x \to c} f(x)\) exists). This means the left-hand limit and the right-hand limit must be equal (\(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\)).
3. The limit must equal the function value (\(\lim_{x \to c} f(x) = f(c)\)).

Let's check these conditions for \(x = 2\):

• Is \(f(2)\) defined? Yes, the first part of the function definition (\(-1 \le x \le 2\)) applies to \(x=2\). \(f(2) = 3(2)^2 + 12(2) - 1 = 3(4) + 24 - 1 = 12 + 24 - 1 = 35\).
• Does the limit as \(x\) approaches 2 exist?
  • Left-hand limit (\(x \to 2^-\)): We use the first part of the function (\(3x^2 + 12x - 1\)) because it applies for \(x \le 2\). \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (3x^2 + 12x - 1) = 3(2)^2 + 12(2) - 1 = 12 + 24 - 1 = 35\).
  • Right-hand limit (\(x \to 2^+\)): We use the second part of the function (\(37 - x\)) because it applies for \(x > 2\). \(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (37 - x) = 37 - 2 = 35\).
  Since the left-hand limit (35) equals the right-hand limit (35), the limit as \(x\) approaches 2 exists and is equal to 35 (\(\lim_{x \to 2} f(x) = 35\)).
• Does the limit equal the function value? We have \(f(2) = 35\) and \(\lim_{x \to 2} f(x) = 35\). Since \(\lim_{x \to 2} f(x) = f(2)\), the third condition is met.

All three conditions for continuity at \(x=2\) are satisfied. Therefore, Statement 1 is correct.

Statement 2: f(x) attains greatest value at x = 2

To find the greatest value (maximum) of the function on its domain \([-1, 3]\), we need to examine the function's behavior in each piece and at the critical point(s) within the intervals, as well as at the endpoints of the entire domain.

Piece 1: \(f(x) = 3x^2 + 12x - 1\) for \(-1 \le x \le 2\).

This is a quadratic function. The vertex of the parabola \(ax^2 + bx + c\) is at \(x = -b/(2a)\). Here, \(a=3\) and \(b=12\). Vertex x-coordinate: \(x = -12/(2 \times 3) = -12/6 = -2\). This vertex is outside the interval \([-1, 2]\). For a parabola opening upwards (\(a > 0\)), the maximum or minimum on a closed interval occurs at the endpoints if the vertex is outside the interval.

Evaluate \(f(x)\) at the endpoints of this interval:

• At \(x = -1\): \(f(-1) = 3(-1)^2 + 12(-1) - 1 = 3 - 12 - 1 = -10\).
• At \(x = 2\): \(f(2) = 3(2)^2 + 12(2) - 1 = 12 + 24 - 1 = 35\).

On the interval \([-1, 2]\), the function value ranges from -10 to 35. The maximum value in this piece is 35, occurring at \(x=2\).

Piece 2: \(f(x) = 37 - x\) for \(2 < x \le 3\).

This is a linear function with a negative slope (-1), meaning it is decreasing on this interval. As \(x\) increases from values slightly greater than 2 up to 3, the function value decreases.

Evaluate \(f(x)\) at the right endpoint of this interval:

• At \(x = 3\): \(f(3) = 37 - 3 = 34\).

As \(x\) approaches 2 from the right (\(x \to 2^+\)), the function value approaches \(37 - 2 = 35\). Since the interval is \(x > 2\), the value 35 is not strictly *attained* within this open part of the interval, but it is the supremum (least upper bound) of the function values on \((2, 3]\).

Now let's compare the values at key points across the entire domain \([-1, 3]\):

• \(f(-1) = -10\)
• \(f(2) = 35\)
• \(f(3) = 34\)

Considering the behavior, the function starts at -10 at \(x=-1\), increases to 35 at \(x=2\), and then decreases from 35 (approaching from the right) to 34 at \(x=3\).

By comparing the values at the endpoints and the point where the definition changes (\(x=2\)), we see that the greatest value attained by the function on the interval \([-1, 3]\) is 35, which occurs at \(x = 2\).

Therefore, Statement 2 is correct.

Statement 3: f(x) is differentiable at x = 2

For a function to be differentiable at a point \(x=c\), it must first be continuous at \(x=c\). We have already established that \(f(x)\) is continuous at \(x=2\).

Next, the left-hand derivative must equal the right-hand derivative at \(x=2\).

First, let's find the derivative of each piece of the function:

• For \(-1 \le x < 2\): \(f'(x) = \frac{d}{dx}(3x^2 + 12x - 1) = 6x + 12\).
• For \(2 < x \le 3\): \(f'(x) = \frac{d}{dx}(37 - x) = -1\).

Now, let's evaluate the left-hand derivative and the right-hand derivative at \(x=2\):

• Left-hand derivative at \(x=2\) (\(x \to 2^-\)): \(\lim_{x \to 2^-} f'(x) = \lim_{x \to 2^-} (6x + 12) = 6(2) + 12 = 12 + 12 = 24\).
• Right-hand derivative at \(x=2\) (\(x \to 2^+\)): \(\lim_{x \to 2^+} f'(x) = \lim_{x \to 2^+} (-1) = -1\).

Since the left-hand derivative at \(x=2\) (24) is not equal to the right-hand derivative at \(x=2\) (-1), the function is not differentiable at \(x=2\).

Therefore, Statement 3 is incorrect.

Summary of Statement Analysis

Let's summarize our findings:

Based on this analysis, statements 1 and 2 are correct, while statement 3 is incorrect.

Final Answer Selection

We are looking for the option that includes only statements 1 and 2. Reviewing the given options:

• Option 1: 1 and 2 only
• Option 2: 2 and 3 only
• Option 3: 1 and 3 only
• Option 4: 1, 2 and 3

The analysis shows that only statements 1 and 2 are correct. This corresponds to Option 1.

Revision Table: Continuity, Differentiability, and Extrema

Additional Information: Piecewise Function Analysis

Analyzing piecewise functions often involves checking their behavior at the points where the definition changes. These points are crucial because properties like continuity and differentiability can fail there, even if the function is smooth everywhere else within its defined pieces.

Steps for analyzing piecewise functions:

1. Continuity: Check the point where the definition changes. Evaluate the left-hand limit, right-hand limit, and function value at that point. If they are all equal, the function is continuous there. Also, ensure each piece is continuous within its defined interval (polynomials and linear functions are generally continuous everywhere).
2. Differentiability: First, check for continuity at the point where the definition changes. If not continuous, it's not differentiable. If continuous, find the derivative of each piece. Evaluate the left-hand derivative and right-hand derivative at the point. If they are equal, the function is differentiable there. Again, check if the derivative of each piece exists within its open interval.
3. Finding Extrema (Greatest/Least Value):
   • Find critical points within each open interval where \(f'(x) = 0\) or \(f'(x)\) is undefined.
   • Evaluate the function \(f(x)\) at these critical points.
   • Evaluate the function \(f(x)\) at the endpoints of the entire domain.
   • Evaluate the function \(f(x)\) at the points where the definition of the function changes (like \(x=2\) in this problem), provided these points are in the domain.
   • Compare all the values found in the previous steps. The largest is the greatest value, and the smallest is the least value over the domain.

In this specific problem, the point of interest \(x=2\) is where the function definition changes, making it a critical point for examining continuity and differentiability. For finding the greatest value, we consider the endpoints \(-1\) and \(3\), and the change point \(2\).