Answer is: option3

Solution:

\( f'(x) > 0 \) for all \( x \in [2, 8] \): So, the function is increasing.
\( f''(x) > 0 \) for all \( x \in [2, 8] \): So, the function is concave up (the slope is increasing).

We need to choose the table of values that reflects both increasing function and increasing slope.

Option (A)

Check if function is increasing:

1. \( f(4) > f(2) \): \( 3 > -1 \) ✅
2. \( f(6) > f(4) \): \( 6 > 3 \) ✅
3. \( f(8) > f(6) \): \( 8 > 6 \) ✅

Check slope (approximate):

1. From 2 to 4: \( \frac{3 - (-1)}{4 - 2} = 2 \)
2. From 4 to 6: \( \frac{6 - 3}{6 - 4} = 1.5 \)
3. From 6 to 8: \( \frac{8 - 6}{8 - 6} = 1 \)

Slope is decreasing → So \( f''(x) < 0 \), not valid

Option (B)

Check increasing:

1. All values are increasing ✅

Slope:

1. 2 to 4: \( \frac{2 - (-1)}{2} = 1.5 \)
2. 4 to 6: \( \frac{5 - 2}{2} = 1.5 \)
3. 6 to 8: \( \frac{8 - 5}{2} = 1.5 \)

Slope is constant → \( f''(x) = 0 \) → So this also violates the condition

Option (C)

Increasing:

1. All values increase ✅

Slopes:

1. 2 to 4: \( \frac{1 - (-1)}{2} = 1 \)
2. 4 to 6: \( \frac{4 - 1}{2} = 1.5 \)
3. 6 to 8: \( \frac{8 - 4}{2} = 2 \)

Slopes are increasing ✅ → So \( f''(x) > 0 \)

This satisfies both conditions!

✅ Correct answer: (C)

Option (D)

Function is decreasing → \( f'(x) < 0 \), violates condition

So, option C