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

2. Quiz Homework Problem: Related Rates 3-10 Gravel 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) 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). P(c,f(c)) a A(a,f(a)) b P1P1 P2P2 a y b B(b,f(b)) c2c2 c1c1 x y x c 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, m tangent = average rate of change, m secant ) b – a or equivalently: f(b) – f(a) = f’(c)(b – a)

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(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 f’(x) = -2x + 6

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

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] 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 a) continuous on the closed interval [a,b]  ok b) differentiable on the open interval (a,b)  ok on open

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

11. 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 a b y x c a b y x c a b y x c A. B. C. 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. a b y xc1c1 c2c2

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] f’(x) = 2x 0 in the interval [-3,3] f’(x) = 0 when x = 0 a) continuous on the closed interval [a,b]  polynomial b)differentiable on the open interval (a,b)  polynomial c)f(a) = f(b)  f(-3) = 18 f(3) = 18

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] f’(x) = 3x² - 4x – 1 -0.2153 and 1.5486 in the interval [-1,2] f’(x) = 0 when x = (  7 + 2)/3 = 1.5486 a) continuous on the closed interval [a,b]  polynomial b)differentiable on the open interval (a,b)  polynomial c)f(a) = f(b)  f(-1) = 0 f(2) = 0 f’(x) = 0 when x = -(  7 - 2)/3 = -0.2153

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 b)differentiable on the open interval (a,b)  no at x = 0 c)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] f’(x) = cos x π/2 in the interval [0,π] f’(x) = 0 (or undefined) when x = π/2 a) continuous on the closed interval [a,b]  ok b)differentiable on the open interval (a,b)  ok c)f(a) = f(b)  f(0) = 0 f(π) = 0

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