Slide 1

  • Topic: Derivatives - Problems on Maximum & Minimum Value of a Function.

Slide 2

  • In calculus, finding the maximum and minimum values of a function is crucial.
  • These points are known as the critical points of the function.
  • We can determine the maximum and minimum values by analyzing the behavior of the derivative.

Slide 3

  • To find the critical points of a given function, we must first calculate the derivative of the function.
  • The critical points occur where the derivative is either zero or undefined.

Slide 4

  • Steps to find the maximum and minimum values:
    1. Find the derivative of the function.
    2. Set the derivative equal to zero and solve for x.
      • These values of x may indicate the presence of maximum or minimum points.
    3. Evaluate the function at these critical points.
      • Determine if the points represent maximum or minimum values.

Slide 5

  • Example: Find the critical points and determine whether they represent maximum or minimum values for the function f(x) = x^3 - 3x^2 - 9x + 3.

Slide 6

  • Step 1: Find the derivative of the function f(x) = x^3 - 3x^2 - 9x + 3.
    • f’(x) = 3x^2 - 6x - 9

Slide 7

  • Step 2: Set the derivative equal to zero and solve for x.
    • 3x^2 - 6x - 9 = 0

Slide 8

  • Solving the equation:
    • 3(x^2 - 2x - 3) = 0
    • (x - 3)(x + 1) = 0

Slide 9

  • The critical points are:
    • x = 3 and x = -1

Slide 10

  • Step 3: Evaluate the function at the critical points.
    • f(3) = 3^3 - 3(3)^2 - 9(3) + 3
    • f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 3

Slide 11

  • Evaluate f(3):
    • f(3) = 3^3 - 3(3)^2 - 9(3) + 3
    • f(3) = 27 - 27 - 27 + 3
    • f(3) = -24
  • Evaluate f(-1):
    • f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 3
    • f(-1) = -1 - 3 + 9 + 3
    • f(-1) = 8

Slide 12

  • Since the function f(x) = x^3 - 3x^2 - 9x + 3 has a maximum point and a minimum point, we need to determine which is which.
  • To do this, we can analyze the behavior of the derivative.

Slide 13

  • Analyzing the behavior of the derivative:
    • f’(x) = 3x^2 - 6x - 9
  • The sign of the derivative can help us determine whether a point is a maximum or minimum.

Slide 14

  • If the derivative is positive, the function is increasing.
    • This means the point is a minimum.

Slide 15

  • If the derivative is negative, the function is decreasing.
    • This means the point is a maximum.

Slide 16

  • Let’s calculate the derivative at the critical points.
  • For x = 3:
    • f’(3) = 3(3)^2 - 6(3) - 9
    • f’(3) = 27 - 18 - 9
    • f’(3) = 0
  • For x = -1:
    • f’(-1) = 3(-1)^2 - 6(-1) - 9
    • f’(-1) = 3 + 6 - 9
    • f’(-1) = 0

Slide 17

  • Since the derivative is zero at both critical points, we cannot determine the nature of these points using the first derivative test.
  • We need to use the second derivative test to make a definitive conclusion.

Slide 18

  • Second derivative test:
    • If f’’(x) < 0, the point represents a maximum.
    • If f’’(x) > 0, the point represents a minimum.
  • Let’s calculate the second derivative.

Slide 19

  • Calculating the second derivative:
    • f’’(x) = 6x - 6
  • Evaluating f’’(3):
    • f’’(3) = 6(3) - 6
    • f’’(3) = 18 - 6
    • f’’(3) = 12
  • Evaluating f’’(-1):
    • f’’(-1) = 6(-1) - 6
    • f’’(-1) = -6 - 6
    • f’’(-1) = -12

Slide 20

  • Conclusion:
    • f’’(3) > 0, so the point (3, -24) represents a minimum.
    • f’’(-1) < 0, so the point (-1, 8) represents a maximum.
  • Therefore, the function f(x) = x^3 - 3x^2 - 9x + 3 has a maximum value of 8 at x = -1 and a minimum value of -24 at x = 3.

Slide 21

  • Let’s work on more examples to solidify our understanding of finding maximum and minimum values.
  • Example: Find the maximum and minimum values of the function f(x) = 2x^3 - 9x^2 + 12x + 7.

Slide 22

  • Step 1: Find the derivative of the function f(x) = 2x^3 - 9x^2 + 12x + 7.
    • f’(x) = 6x^2 - 18x + 12

Slide 23

  • Step 2: Set the derivative equal to zero and solve for x.
    • 6x^2 - 18x + 12 = 0
    • Simplifying the equation, we get:
    • x^2 - 3x + 2 = 0
    • Factoring the equation, we have:
    • (x - 1)(x - 2) = 0

Slide 24

  • The critical points are:
    • x = 1 and x = 2

Slide 25

  • Step 3: Evaluate the function at the critical points.
    • f(1) = 2(1)^3 - 9(1)^2 + 12(1) +7
    • f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 7

Slide 26

  • Evaluating f(1):
    • f(1) = 2(1)^3 - 9(1)^2 + 12(1) +7
    • f(1) = 2 - 9 + 12 + 7
    • f(1) = 12
  • Evaluating f(2):
    • f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 7
    • f(2) = 16 - 36 + 24 + 7
    • f(2) = 11

Slide 27

  • Analyzing the behavior of the derivative:
    • f’(x) = 6x^2 - 18x + 12
  • Let’s calculate the derivative at the critical points.

Slide 28

  • For x = 1:
    • f’(1) = 6(1)^2 - 18(1) + 12
    • f’(1) = 6 - 18 + 12
    • f’(1) = 0
  • For x = 2:
    • f’(2) = 6(2)^2 - 18(2) + 12
    • f’(2) = 24 - 36 + 12
    • f’(2) = 0

Slide 29

  • Since the derivative is zero at both critical points, we cannot determine the nature of these points using the first derivative test.
  • Let’s use the second derivative test.

Slide 30

  • Calculating the second derivative:
    • f’’(x) = 12x - 18
  • Evaluating f’’(1):
    • f’’(1) = 12(1) - 18
    • f’’(1) = 12 - 18
    • f’’(1) = -6
  • Evaluating f’’(2):
    • f’’(2) = 12(2) - 18
    • f’’(2) = 24 - 18
    • f’’(2) = 6
  • Therefore, the function f(x) = 2x^3 - 9x^2 + 12x + 7 has a maximum value of 12 at x = 1 and a minimum value of 11 at x = 2.