1 / 16

Lesson 4-2

Lesson 4-2. Mean Value Theorem and Rolle’s Theorem. Quiz.

abram
Download Presentation

Lesson 4-2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lesson 4-2 Mean Value Theorem and Rolle’s Theorem

  2. Quiz • Homework Problem: Related Rates 3-10Gravel is being dumped from a conveyor belt at a rate of 30 ft³/min, and forms a pile in shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10ft high? • Reading questions: • What were the names of the two theorems in section 4.2? • What pre-conditions (hypotheses) do the Theorems have in common?

  3. Objectives • Understand Rolle’s Theorem • Understand Mean Value Theorem

  4. Vocabulary • Existence Theorem – a theorem that guarantees that there exists a number with a certain property, but it doesn’t tell us how to find it.

  5. Theorems Mean Value Theorem: Let f be a function that is a) continuous on the closed interval [a,b] b) differentiable on the open interval (a,b) then there is a number c in (a,b) such that f(b) – f(a) f’(c) = --------------- or equivalently: f(b) – f(a) = f’(c)(b – a) b – a Rolle’s Theorem: Let f be a function that is a) continuous on the closed interval [a,b] b) differentiable on the open interval (a,b) c) f(a) = f(b) then there is a number c in (a,b) such that f’(c) = 0

  6. Mean Value Theorem (MVT) Let f be a function that is a) continuous on the closed interval [a,b] b) differentiable on the open interval (a,b) then there is a number c in (a,b) such that f(b) – f(a) f’(c) = --------------- (instantaneous rate of change, mtangent = average rate of change, msecant) b – a or equivalently: f(b) – f(a) = f’(c)(b – a) P(c,f(c)) y y P1 A(a,f(a)) P2 B(b,f(b)) x x c2 c1 a a c b b For a differentiable function f(x), the slope of the secant line through (a, f(a)) and (b, f(b)) equals the slope of the tangent line at some point c between a and b. In other words, the average rate of change of f(x) over [a,b] equals the instantaneous rate of change at some point c in (a,b).

  7. Example 1 Verify that the mean value theorem (MVT) holds for f(x) = -x² + 6x – 6 on [1,3]. a) continuous on the closed interval [a,b]  polynomial b) differentiable on the open interval (a,b)  polynomial f’(x) = -2x + 6 f(b) – f(a) = f(3) – f(1) = 3 – (-1) = 4 f(b) – f(a) / (b – a) = 4/2 = 2 f’(x) = 2 = 6 – 2x so -4 = -2x 2 = x

  8. Example 2 Find the number that satisfies the MVT on the given interval or state why the theorem does not apply. f(x) = x2/5 on [0,32] a) continuous on the closed interval [a,b]  ok b) differentiable on the open interval (a,b)  ok on open f’(x) = (2/5)x-3/5 f(b) – f(a) = f(32) – f(0) = 4 – 0 = 4 f(b) – f(a) / (b – a) = 4/32 = 0.125 f’(x) = 0.125 = (2/5)x-3/5 so x = 6.94891

  9. Example 3 Find the number that satisfies the MVT on the given interval or state why the theorem does not apply. f(x) = x + (1/x) on [1,3] a) continuous on the closed interval [a,b]  ok b) differentiable on the open interval (a,b)  ok on open f’(x) = 1 – x-2 f(b) – f(a) = f(3) – f(1) = 10/3 – 2 = 4/3 f(b) – f(a) / (b – a) = (7/3)/2 = 2/3 f’(x) = 2/3 = 1 – x-2 so x-2 = 1/3 x² = 3 x=3 = 1.732

  10. Example 4 Find the number that satisfies the MVT on the given interval or state why the theorem does not apply. f(x) = x1/2 + 2(x – 3)1/3 on [1,9] a) continuous on the closed interval [a,b]  ok b) differentiable on the open interval (a,b)  vertical tan f’(x) = 1/2x-1/2 + 2/3(x – 3)-2/3 f’(x) undefined at x = 3 (vertical tangent) MVT does not apply

  11. Rolle’s Theorem y y y y c x x c c1 c2 x x c a a a a b b b b Let f be a function that is a) continuous on the closed interval [a,b] b) differentiable on the open interval (a,b) c) f(a) = f(b) then there is a number c in (a,b) such that f’(c) = 0 A. C. B. D. Case 1: f(x) = k (constant) [picture A] Case 2: f(x) > f(a) for some x in (a,b) [picture B or C] Extreme value theorem guarantees a max value somewhere in [a,b]. Since f(a) = f(b), then at some c in (a,b) this max must occur. Fermat’s Theorem says that f’(c) =0. Case 3: f(x) < f(a) for some x in (a,b) [picture C or D] Extreme value theorem guarantees a min value somewhere in [a,b]. Since f(a) = f(b), then at some c in (a,b) this min must occur. Fermat’s Theorem says that f’(c) =0.

  12. Example 5 Determine whether Rolle’s Theorem’s hypotheses are satisfied &, if so, find a number c for which f’(c) = 0. f(x) = x² + 9 on [-3,3] • a) continuous on the closed interval [a,b]  polynomial • differentiable on the open interval (a,b)  polynomial • f(a) = f(b)  f(-3) = 18 f(3) = 18 f’(x) = 2x f’(x) = 0 when x = 0 0 in the interval [-3,3]

  13. Example 6 Determine whether Rolle’s Theorem’s hypotheses are satisfied &, if so, find a number c for which f’(c) = 0. f(x) = x³ - 2x² - x + 2 on [-1,2] • a) continuous on the closed interval [a,b]  polynomial • differentiable on the open interval (a,b)  polynomial • f(a) = f(b)  f(-1) = 0 f(2) = 0 f’(x) = 3x² - 4x – 1 f’(x) = 0 when x = (7 + 2)/3 = 1.5486 f’(x) = 0 when x = -(7 - 2)/3 = -0.2153 -0.2153 and 1.5486 in the interval [-1,2]

  14. Example 7 Determine whether Rolle’s Theorem’s hypotheses are satisfied &, if so, find a number c for which f’(c) = 0. f(x) = (x² - 1) / x on [-1,1] • a) continuous on the closed interval [a,b]  no at x = 0 • differentiable on the open interval (a,b)  no at x = 0 • f(a) = f(b)  f(-1) = 0 f(1) = 0

  15. Example 8 Determine whether Rolle’s Theorem’s hypotheses are satisfied &, if so, find a number c for which f’(c) = 0. f(x) = sin x on [0,π] • a) continuous on the closed interval [a,b]  ok • differentiable on the open interval (a,b)  ok • f(a) = f(b)  f(0) = 0 f(π) = 0 f’(x) = cos x f’(x) = 0 (or undefined) when x = π/2 π/2 in the interval [0,π]

  16. Summary & Homework • Summary: • Mean Value and Rolle’s theorems are existance theorems • Each has some preconditions that must be met to be used • Homework: • pg 295-296: 2, 7, 11, 12, 24

More Related