Let's solve the problems from the image.

Part (a) - $g(x) = 4x - 10$

The average rate of change of a function $f(x)$ over an interval $[x_1, x_2]$ is given by the formula:

$\text{Average rate of change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

(i) Interval: $[3, 5]$

For $g(x) = 4x - 10$:

1. Calculate $g(5)$: $g(5) = 4(5) - 10 = 20 - 10 = 10$

2. Calculate $g(3)$: $g(3) = 4(3) - 10 = 12 - 10 = 2$

3. Average rate of change: $\frac{g(5) - g(3)}{5 - 3} = \frac{10 - 2}{2} = \frac{8}{2} = 4$

(ii) Interval: $[-7, -4]$

1. Calculate $g(-4)$: $g(-4) = 4(-4) - 10 = -16 - 10 = -26$

2. Calculate $g(-7)$: $g(-7) = 4(-7) - 10 = -28 - 10 = -38$

3. Average rate of change: $\frac{g(-4) - g(-7)}{-4 - (-7)} = \frac{-26 - (-38)}{-4 + 7} = \frac{-26 + 38}{3} = \frac{12}{3} = 4$

Part (b) - $f(x) = 2x^3 - x^2 + 1$

(i) Interval: $[3, 5]$

1. Calculate $f(5)$: $f(5) = 2(5)^3 - (5)^2 + 1 = 2(125) - 25 + 1 = 250 - 25 + 1 = 226$

2. Calculate $f(3)$: $f(3) = 2(3)^3 - (3)^2 + 1 = 2(27) - 9 + 1 = 54 - 9 + 1 = 46$

3. Average rate of change: $\frac{f(5) - f(3)}{5 - 3} = \frac{226 - 46}{2} = \frac{180}{2} = 90$

(ii) Interval: $[-7, -4]$

1. Calculate $f(-4)$: $f(-4) = 2(-4)^3 - (-4)^2 + 1 = 2(-64) - 16 + 1 = -128 - 16 + 1 = -143$

2. Calculate $f(-7)$: $f(-7) = 2(-7)^3 - (-7)^2 + 1 = 2(-343) - 49 + 1 = -686 - 49 + 1 = -734$

3. Average rate of change: $\frac{f(-4) - f(-7)}{-4 - (-7)} = \frac{-143 - (-734)}{-4 + 7} = \frac{-143 + 734}{3} = \frac{591}{3} = 197$

Final Answers:

• Part (a):

  • (i) 4
  • (ii) 4

• Part (b):

  • (i) 90
  • (ii) 197

Would you like more details on any of these steps or have questions?

Here are 5 related questions:

1. What is the geometric meaning of the average rate of change?
2. How does the average rate of change compare to the derivative?
3. Can we apply the same process to trigonometric functions?
4. How do you interpret negative rates of change?
5. Can you generalize the formula for average rate of change?

Tip: The average rate of change of a linear function is constant across all intervals.