Answer is: option2

I and III only

Solution:

The average rate of change of a function \( f(x) \) over the interval \([a, b]\) is given by:

\[ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} \]

From the graph:

• \( f(-3) = 0 \)
• \( f(6) = -1 \)

So, the average rate of change over \([-3,6]\) is:

\[ \frac{f(6) - f(-3)}{6 - (-3)} = \frac{-1 - 0}{6 + 3} = \frac{-1}{9} \]

This matches Statement I, so it is true.

The quarter-circle sections have positive slopes.

The line segment from (2,3) to (6,-1) has slope:

\[ \frac{-1 - 3}{6 - 2} = \frac{-4}{4} = -1 \]

which is much steeper than \(-\frac{1}{9}\).

Since nowhere on the differentiable sections does the slope equal \(-\frac{1}{9}\), there is no point \( c \) where \( f'(c) = -\frac{1}{9} \).

Statement II is false because no such \( c \) exists.

Given:

\[ h(x) = f\left(\frac{1}{2} x\right) \]

Using the chain rule:

\[ h'(x) = f'\left(\frac{1}{2}x\right) \cdot \frac{d}{dx}\left(\frac{1}{2}x\right) = f'\left(\frac{1}{2}x\right) \cdot \frac{1}{2} \]

Evaluating at \( x = 6 \):

\[ h'(6) = f'\left(\frac{1}{2} \times 6\right) \cdot \frac{1}{2} = f'(3) \cdot \frac{1}{2} \]

From the graph, at \( x = 3 \), the function is decreasing at a slope of -1, so \( f'(3) = -1 \). Therefore:

\[ h'(6) = (-1) \times \frac{1}{2} = -\frac{1}{2} \]

This matches Statement III, so it is true.

Thus, the correct answer is (B) I and III only.