If f(x)=`{{:(,x-5 "for "x le 1),(,4x^(2)-9"for "1 lt x lt 2"then "f'(2+)),(,3x+4"for "xge2):}`

Options

A. 0
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find the right-hand derivative of the function \( f(x) \) at \( x = 2 \). The function is defined piecewise as follows:

1. \( f(x) = x - 5 \) for \( x \leq 1 \)
2. \( f(x) = 4x^2 - 9 \) for \( 1 < x < 2 \)
3. \( f(x) = 3x + 4 \) for \( x \geq 2 \)

### Step-by-Step Solution:

**Step 1: Identify the relevant piece of the function for \( x = 2 \)**

Since we are looking for \( f'(2^+) \), we need to use the piece of the function that applies when \( x \) is greater than or equal to 2. Therefore, we have:

\[ f(x) = 3x + 4 \quad \text{for } x \geq 2 \]

**Step 2: Calculate \( f(2) \)**

We need to find the value of \( f(2) \):

\[ f(2) = 3(2) + 4 = 6 + 4 = 10 \]

**Step 3: Set up the limit for the derivative**

The right-hand derivative \( f'(2^+) \) is defined as:

\[ f'(2^+) = \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} \]

Substituting the expression for \( f(x) \) when \( x \) is greater than or equal to 2:

\[ f'(2^+) = \lim_{x \to 2^+} \frac{(3x + 4) - 10}{x - 2} \]

**Step 4: Simplify the expression**

Now we simplify the expression inside the limit:

\[ f'(2^+) = \lim_{x \to 2^+} \frac{3x + 4 - 10}{x - 2} = \lim_{x \to 2^+} \frac{3x - 6}{x - 2} \]

This can be factored as:

\[ f'(2^+) = \lim_{x \to 2^+} \frac{3(x - 2)}{x - 2} \]

**Step 5: Cancel the common terms**

Since \( x \) approaches \( 2 \) from the right, we can cancel \( (x - 2) \) from the numerator and denominator:

\[ f'(2^+) = \lim_{x \to 2^+} 3 = 3 \]

Thus, we find that:

\[ f'(2^+) = 3 \]

### Final Answer:

The value of \( f'(2^+) \) is \( 3 \).

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