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Extreme Values of Functions. Lesson C.2. Objectives. Students will be able to determine the local or global extreme values of a function. Definition: Absolute (Global) Extreme Values.
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Extreme Values of Functions Lesson C.2
Objectives • Students will be able to • determine the local or global extreme values of a function.
Definition: Absolute (Global) Extreme Values Given a function f with domain D, then f (c) is the absolute maximum value of the function if and only if f (x) <f (c) for all x in the domain. Given a function f with domain D, then f (c) is the absolute minimum value of the function if and only if f (x) >f (c) for all x in the domain.
Examples Absolute Maximum Only Absolute Minimum Only
Exploring Extreme Values • On [ -] • f(x) = sinx • Takes on maximum value of 1 and minimum value of -1
Exploring Extreme Values • On [ -] • f(x) = cosx • Takes on maximum value of 1(once) and minimum value of 0 (twice)
Exploring Extreme Values • y = x² Domain: (-∞, ∞) • No absolute max • Absolute Minimum at (0,0)
Exploring Extreme Values • y = x² Domain: [0,2] Absolute Maximum of 4 at x=2 Absolut Minimum of 0 at x = 0
Theorem 1: The Extreme Value Theorem If f is a continuous function on a closed interval [a, b], then f has both a maximum value and a minimum value on the interval.
Example of E.V.T • The function is continuous over a closed interval [a,b] • Max @ x = b • Min @ x = c2 a c1 c2 b
Decide whether each point is an absolute max/min, relative max/min, or neither.
Definition: Local (Relative) Extreme Values Let c be an interior point on the domain of function f. Then f (c) is a local maximum value at c if and only if f(x) <f(c) for all x in the open interval containing c. f (c) is a local minimum value at c if and only if f(x) >f(c) for all x in the open interval containing c.
Theorem 2: Local Extreme Values Theorem If a function f has a local maximum value or a local minimum value at an interior point c of its domain, and if f ’ exists at c, then f ’(c) = 0.
Definition: Critical Point A critical point at x = c on the interior of the domain of the function f exists where f ’(c) = 0 or f ’(c) is undefined. (Your book refers to this as a critical number) If f has a relative minimum or maximum at x = c, then c is a critical point (number) of f.
Finding Absolute Extrema • Find derivative of function • Find all critical points of f(x) that are in the interval [a,b], you do this by setting derivative equal to 0. • Evaluate the original function at the critical points and at end points • Identify the absolute extrema
Example 1 • Determine the absolute extrema for g(t) = 2t³ + 3t² - 12 t + 4 on [-4,2] Get the derivative: g’(t) = 6t² + 6t – 12 = 0 So we have two critical points at t = -2 and t = 1 Both points lie in the interval so we will use both of them.
Original functiong(t) = 2t³ + 3t² - 12 t + 4 • g(-2) = 24 • g(1) = -3 • g(-4) = -28 • g(2) = 8 • Our absolute max of g(t) is 24 and occurs at t= -2 • Our absolute min of g(t) is -28 and occurs at t = -4
Example 2 • Find the extrema of f(x) = 3x⁴ - 4x³ on the interval [-1, 2]. • Differentiate first: • f’(x) = 12x³ - 12x² • 12x³ - 12x² = 0 at • CP: x = 0 and x = 1
Critical numbers x = 0 and 1 • Evaluate at the critical numbers and endpoints. f(x) = 3x⁴ - 4x³ • F(-1) = 7 • F(0) = 0 • F(1) = -1 • F(2) = 16 Minimum Maximum
Example 3 • Find the extrema of f(x) = 2sinx – cos2x on the interval [0, 2𝝿]. • F’(x) = 2cosx + 2sin2x • 2cosx + 2sin2x = 0 when x = , , ,
Critical Numbers and endpoints f(x) = 2sinx – cos2x • F(0) = -1 • F() = 3 • F() = - 3/2 • F() = -1 • F() = -3/2 • F(2𝝿) = -1 Maximum Minimum minimum
Example 4 • Locate the absolute extrema of the function on the closed interval [0,5]. • F(x) = 2x + 5 • F’(x) = 2 • No critical numbers • F(0) = 5 • F(5) = 15 Minimum Maximum
Example 5 Find the absolute maximum/minimum, local maximum/minimum, and all other critical points. No absolute max/min. (look at the graph!) Local Maximum at (–1, 15); Local Minimum at (2, –12).
Homework C.2 • Pg. 169 • # 2, 19, 21, 23, 29, 36, 53