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Types of Stationary Points .2 Geometrical Application of Calc 1. Minimum - - - - - - -0 + + + + + + + x LHS Minimum RHS f’(x) < 0 = 0 > 0

Types of Stationary Points 2.2 Geometrical Application of Calculus 1. Minimum - - - - - - - 0 + + + + + + + xLHSMinimumRHS f’(x)< 0= 0> 0

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Page 1: Types of Stationary Points 2.2 Geometrical Application of Calculus 1. Minimum - - - - - - - 0 + + + + + + + xLHSMinimumRHS f’(x)< 0= 0> 0

Types of Stationary Points

2.2 Geometrical Application of Calculus

1. Minimum

---- - - - 0 ++

++

+++

x LHS Minimum RHS

f’(x) < 0 = 0 > 0

Page 2: Types of Stationary Points 2.2 Geometrical Application of Calculus 1. Minimum - - - - - - - 0 + + + + + + + xLHSMinimumRHS f’(x)< 0= 0> 0

Types of Stationary Points

2.2 Geometrical Application of Calculus

2. Maximum

x LHS Maximum RHS

f’(x) > 0 = 0 < 0

++

+++ 0 - - ---

-

Page 3: Types of Stationary Points 2.2 Geometrical Application of Calculus 1. Minimum - - - - - - - 0 + + + + + + + xLHSMinimumRHS f’(x)< 0= 0> 0

Types of Stationary Points

2.2 Geometrical Application of Calculus

3.Point ofHorizontalInflection

x LHS Inflection RHS

f’(x) > 0 0 > 0

f’(x) < 0 0 < 0

+++

+ 0 ++++ ----- - 0 - ----

Page 4: Types of Stationary Points 2.2 Geometrical Application of Calculus 1. Minimum - - - - - - - 0 + + + + + + + xLHSMinimumRHS f’(x)< 0= 0> 0

Types of Stationary Points

2.2 Geometrical Application of Calculus

1. Find any stationary points on the curve f(x) = x2 - 2x & determine what type it is.f’(x) = 2x - 2

2x - 2 = 0 (Stationary)2x = 2

f(1) = (1)2 – 2(1) = -1

LHS x = 0 Stationary x = 1 RHS x = 2

- 0 +

f’(0) = 2(0) - 2 = -2f’(2) = 2(2) - 2 = +2

(1, -1) is a Minimum

Stationary @ (1, -1)

x = 1

Page 5: Types of Stationary Points 2.2 Geometrical Application of Calculus 1. Minimum - - - - - - - 0 + + + + + + + xLHSMinimumRHS f’(x)< 0= 0> 0

Types of Stationary Points

2.2 Geometrical Application of Calculus

2. Find the turning point on the curve y = 2x3 + 2 and determine what type it is.

f(x) = 2x3 + 2f’(x) = 6x2

6x2 = 0 (Stationary)x = 0

f(0) = 2(0)3 + 2 = 2

LHS x = -1 Stationary x = 0 RHS x = 1

+ 0 +

f’(-1) = 6(-1)2 = +6

f’(1) = 6(1)2 = +6

(0, 2) is a horizontal point of inflection

Stationary @ (0, 2)