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MA117 Differentiation Study Notes

1. Average Gradient

• Definition: The average gradient is defined as the gradient of a straight line that connects two points on a curve.
  • It is used to calculate the rate of change between two variables: the x-coordinate and the y-coordinate on a graph.
  • Mathematically expressed as:
    $ext{Average Gradient} = \frac{Y_2 - Y_1}{X_2 - X_1}$

2. Calculating the Average Gradient

• Example Calculation: The gradient of the line connecting points A and C is calculated via the formula:
  • For example, using points A and C with coordinates (A(-1, -1)) and (C(7, 5)):
  • Coordinates: (Y_2 = 7), (Y_1 = -1), (X_2 = 5), (X_1 = -1)
  • Calculation:
    $ext{Gradient} = \frac{Y_C - Y_A}{X_C - X_A} = \frac{5 - (-1)}{7 - (-1)} = \frac{6}{8} = -4$

3. Finding the Average Gradient Between Two Points

• Step-by-Step Process:

  1. Assign labels to the x-values: (x = a) and (x = a + h).
  2. Calculate corresponding y-coordinates using the function (g(x) = x^2).
  • Thus, (y_1 = g(a) = a^2) and (y_2 = g(a+h) = (a+h)^2).
  1. Plug coordinates into the average gradient formula:

  • Calculating:\
    $y_2 - y_1 = (a+h)^2 - a^2 = a^2 + 2ah + h^2 - a^2 = 2ah + h^2$
    $x_2 - x_1 = (a+h) - a = h$

    • Therefore,
      $ext{Average Gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2ah + h^2}{h} = 2a + h$

4. Effect of Changing Point Positions on Average Gradient

• Observation: If point a moves closer to point P, h becomes smaller. When point P and point A overlap, h approaches zero:
  • At h = 0, the average gradient becomes:
    $ext{Gradient} = 2a$

5. Example Problem on Average Gradient

• Problem: Find the average gradient between points P (2, g(2)) and Q (5, g(5)).
  • Solution involves recognizing that (a = 2), ( a + h = 5) implies (h = 3).
  • The average gradient is:
    $ext{Average Gradient} = 2a = 2(2) + 3 = 7$

6. Quadratic Functions

• Consider points A (-4, 5) and B (4, 5).
  • Calculation steps:
  • Average Gradient Formula: $ext{Average Gradient} = \frac{Y_B - Y_A}{X_B - X_A}$
  • Substitute coordinates:
    $ext{Average Gradient} = \frac{5 - 5}{4 - (-4)} = \frac{0}{8} = 0$
  • Conclusion: The average gradient between points A and B is 0.

7. Hyperbola and Linear Functions Example

• Given points W (-3, 3) and Z (-1, 7) for a hyperbola and linear function:
  • Use the average gradient formula:
  • Calculate for points W and Z:
    $ext{Average Gradient} = \frac{7 - 3}{-1 - (-3)} = \frac{4}{2} = 2$
  • The average gradient for the linear function is also 2, confirming the average gradient between W and Z is equal.

8. Difference Between Gradient and Average Gradient

• Gradient: Refers to the slope of the tangent at a single point on a function.
• Average Gradient: Refers to the slope of the secant line connecting two points on the curve.

9. Additional Examples and Exercises

• Example 1:

  • Given function: (f(x) = x^2 - 3x + 2)
  • For (x = 1) and (x = 4):\
  1. Calculate (f(1)) and (f(4)):
     • (f(1) = (1)^2 - 3(1) + 2 = 0)
     • (f(4) = (4)^2 - 3(4) + 2 = 6)
  2. Apply average gradient formula:
     $ext{Average Gradient} = \frac{6 - 0}{4 - 1} = \frac{6}{3} = 2$

• Example 2:

  • Given function: (g(x) = 2 ext{e}^x) and calculate average gradient between (x = -1) and (x = -4).
  • Evaluate function values:
  1. Find (g(-1)) and (g(-4)) respectively to derive average gradient.

Conclusion

• Understanding the roles and calculation methods of both average and instantaneous (point-based) gradients is crucial in differentiation. Students should practice various examples for mastery, focusing on both the quantitative relationships defined by the gradient formulas and the qualitative implications of these mathematical concepts in calculus.