calculus date: 12/10/13 obj: swbat apply rolle’s and the mean value thm //youtu.be/4j2zltgoile...
Post on 17-Jan-2016
215 Views
Preview:
TRANSCRIPT
• Calculus Date: 12/10/13
Obj: SWBAT apply Rolle’s and the Mean Value Thmhttp://youtu.be/4j2ZLtGoiLE derivatives of absolute values
http://youtu.be/PBKnttVMbV4 first derivative test inc. dec.
Do Now –
HW Requests: pg 198 #11, 12, 13, 15, 17, 22, 23, 25, 29, 39, 41
SM pg 103 Read Section 5.2• pg 183 #1-21 odds• In class: page SM 82/83• HW: pg SM 84/85• Step by Step Manual pg 4, 5, 8, 11-Friday• In class: SM 102• Announcements:• No class next Wednesday• Step by Step Assignment Friday
"Do not judge me by my successes, judge me by how many times
I fell down and got back up again.“Nelson Mandela
The Mean Value Theorem
Lesson 4.2
I wonder how mean this
theorem really is?
Think About It
• Consider a trip of two hours that is 120 miles in distance … You have averaged 60 miles per hour
• What reading on your speedometer would you have expected to see at least once?
60
Rolle’s Theorem
• Given f(x) on closed interval [a, b] Differentiable on open interval (a, b)
• If f(a) = f(b) … then There exists at least one number
a < c < b such that f ’(c) = 0
f(a) = f(b)
a bc
Ex. Find the two x-intercepts of f(x) = x2 – 3x + 2and show that f’(x) = 0 at some point between the
two intercepts.
f(x) = x2 – 3x + 20 = (x – 2)(x – 1)x-int. are 1 and 2
f’(x) = 2x - 3
0 = 2x - 3
x = 3/2
Rolles Theorem is satisfied as there is a point atx = 3/2 where f’(x) = 0.
Let f(x) = x4 – 2x2 . Find all c in the interval (-2, 2)such that f’(x) = 0.
Since f(-2) and f(2) = 8, we can use Rolle’s Theorem.
f’(x) = 4x3 – 4x = 0
4x(x2 – 1) = 0
x = -1, 0, and 1
Thus, in the interval(-2, 2), the derivative
is zero at each of thesethree x-values.
8
Mean Value Theorem• We can “tilt” the picture of Rolle’s Theorem
Stipulating that f(a) ≠ f(b)
• Then there exists a c such that
a bc
( ) ( )'( )
f b f af c
b a
The Mean Value Theorem (MVT)aka the ‘crooked’ Rolle’s Theorem
If f is continuous on [a, b] and differentiable on (a, b)
There is at least one number c on (a, b) at which
ab
f(a)
f(b)
c
Conclusion:Slope of Secant Line
EqualsSlope of Tangent Line
ab
afbfcf
)()(
)('
We can “tilt” the picture of Rolle’s TheoremStipulating that f(a) ≠ f(b)
Finding c
• Given a function f(x) = 2x3 – x2 Find all points on the interval [0, 2] where
• Strategy Find slope of line from f(0) to f(2) Find f ‘(x) Set f ‘(x) equal to slope … solve for x
( ) ( )'( )
f b f af c
b a
Given f(x) = 5 – 4/x, find all c in the interval (1,4)such that the slope of the secant line = the slope of
the tangent line.
?
14
)1()4()('
ff
cf 114
14
?)(' xf2
4
x
14
2
x24 x2x
But in the interval of (1,4),only 2 works, so c = 2.
Find the value(s) of c that satisfy the Mean Value Theorem for
1f x x on 4, 4
x
1 17 1 17f 4 4 f 4 4
4 4 4 4
17 17f b f a 174 4
b a 4 4 16
2
17 1f ' c 1
16 x
2 f b f aIf f x x 2x 1, a 0, b 1, and f ' c , find c.
b a
f(0) = -1 f(1) = 2
f b f a 2 13
b a 1 0
f ' x 2x 2
3 2x 2 1
x2
f(3) = 39 f(-2) = 64
f b f a 64 395
b a 2 3
For how many value(s) of c is f ‘ (c ) = -5?
If , how many numbers on [-2, 3] satisfythe conclusion of the Mean Value Theorem.
2 2f x x 12 x 4
A. 0 B. 1 C. 2 D. 3 E. 4
CALCULATOR REQUIRED
X X X
Find the value(s) of c that satisfy the Mean Value Theorem for
1f x x on 4, 4
x
Note: The Mean Value Theorem requires the function to be continuous on [-4, 4] and differentiable on (-4, 4). Therefore, sincef(x) is discontinuous at x = 0 which is on [-4, 4], there may be no
value of c which satisfies the Mean Value Theorem
Since has no real solution, there is no value of c on
[-4, 4] which satisfies the Mean Value Theorem
2
1 1
16 x
Mean Value Theorem
• Applied to a cubic equation
Note Geogebera Example
Note Geogebera Example
Find the value(s) of c which satisfy Rolle’s Theorem for on the interval [0, 1]. 4f x x x
Verify…..f(0) = 0 – 0 = 0 f(1) = 1 – 1 = 0
3f ' x 4x 1 30 4x 1
31
c4
which is on [0, 1]
Given the graph of f(x) below, use the graph of f to estimate thenumbers on [0, 3.5] which satisfy the conclusion of the Mean Value
Theorem.
2Determine whether f x x 2x 2 satisfies the hypothesis of
the Mean Value Theorem on -2, 2 . If it does, find all numbers
f b f ac in (a, b) such that f ' c
b a
f(x) is continuous and differentiable on [-2, 2]
f 2 f 2 6 2
2 2 42
f ' x 2x 2
2x 2 2 c 0
On the interval [-2, 2], c = 0 satisfies the conclusion of MVT
2x 1Determine whether f x satisfies the hypothesis of
x 2the Mean Value Theorem on -2, 1 . If it does, find all numbers
f b f ac in (a, b) such that f ' c
b a
f(x) is continuous and differentiable on [-2, 1]
30
f 1 f 2 41 2 3
1
4
2
2
2x x 2 1 x 1
xf ' x
2
2
2
x 4x 1 1
x 4x 4 4
2 24x 16x 4 x 4x 4 23x 12x 0
3x x 4 0 On the interval [-2, 1], c = 0 satisfies the conclusion of MVT
2x 1Determine whether f x satisfies the hypothesis of
x 2the Mean Value Theorem on 0, 4 . If it does, find all numbers
f b f ac in (a, b) such that f ' c
b a
Since f(x) is discontinuous at x = 2, which is part of the interval[0, 4], the Mean Value Theorem does not apply
3Determine whether f x x 3x 1 satisfies the hypothesis of
the Mean Value Theorem on -1, 2 . If it does, find all numbers
f b f ac in (a, b) such that f ' c
b a
f(x) is continuous and differentiable on [-1, 2]
f0
2 f 1 3 3
2 1 3
23x 3f x' 23x 3 0
c = 1 satisfies the conclusion of MVT
3 x 1 x 1 0
Modeling Problem• Two police cars are located at fixed points 6
miles apart on a long straight road. The speed limit is 55 mph A car passes the first point at 53 mph Five minutes later he passes the second at 48
mph Yuk! Yuk! I think he was
speeding, EnosWe need to
prove it, Rosco
top related