Transcript Theorem 3.3 Rolle's Theorem and Figure 3.8

CHAPTER 3
SECTION 3.2
ROLLE’S THEOREM
AND
THE MEAN VALUE THEOREM
Theorem 3.3 Rolle's Theorem and
Figure 3.8
Rolle’s Theorem for Derivatives
Example: Determine whether Rolle’s Theorem can be
applied to f(x) = (x - 3)(x + 1)2 on [-1,3]. Find all
values of c such that f ′(c )= 0.
f(-1)= f(3) = 0 AND f is continuous on [-1,3] and diff on
(1,3) therefore Rolle’s Theorem applies.
f ′(x )= (x-3)(2)(x+1)+ (x+1)2
Multiply and Factor
f ′(x )= (x+1)(3x-5) , set = 0
c = -1 ( not interior on the interval) or 5/3
c = 5/3
Apply Rolle's Theorem
Apply Rolle's Theorem to the following function f and
compute the location c.
f ( x)  x 3  x
on [0, 1]
f ( x)  3x 2  1
f (0)  f (1)  0
By Rolle ' s Theorem there is a c in [0, 1] such that
f (c)  3c 2  1  0
3c 2  1  0
3c 2  1
c2 
c 
1
3
1
3
, [ 13 ]
Theorem 3.4 The Mean Value
Theorem and
Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b   f  a 
ba
 f c
Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b   f  a 
ba
 f c
Differentiable implies that the function is also continuous.
Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b   f  a 
ba
 f c
Differentiable implies that the function is also continuous.
The Mean Value Theorem only applies over a closed interval.

Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then
at some point between a and b:
f b   f  a 
ba
 f c
The Mean Value Theorem says that at some point
in the closed interval, the actual slope equals the
average slope.

Tangent parallel
to chord.
y
Slope of tangent:
f  c
B
Slope of chord:
f b   f  a 
ba
A
0
y  f  x
a
c
b
x

Mean Value Theorem
If f is continuous on [a,b] and differentiable on (a,b) then there exists a
value, c, in (a,b) such that
f ' c  
4
2
a
b
-5
5
-2
-4
f  b   f a
ba
Mean Value Theorem
If f is continuous on [a,b] and differentiable on (a,b) then there exists a
value, c, in (a,b) such that
f ' c  
c can’t be an endpoint
Slope of a
tangent line

f  b   f a
ba
Slope of the line
through the endpoints
4
Instantaneous
rate of change
2
a
c2
-5 c
1
c4
c3
-2
-4
b
5

Average rate of change
1. Apply the MVT to
f  x   x 2  4
on [-1,4].
f  x   x 2  4
1. Apply the MVT to
f(x) is continuous on [-1,4].
f '  x   2x
f(x) is differentiable on [-1,4].
2c 
f  4   f  1
4  1
15
2c 
5
2c  3
3
c
2
on [-1,4].
MVT applies!
2. Apply the MVT to f  x   x
2
3
on [-1,2].
2. Apply the MVT to f  x   x
2
3
on [-1,2].
f(x) is continuous on [-1,2].
f 'x  x
2
 1
3x 3
2
3
 13
f(x) is not differentiable at x = 0.
MVT does not apply!
Alternate form of
the Mean Value Theorem for Derivatives
f (b)  f (a)  (b  a) f '(c)
Determine if the mean value theorem applies, and if
so find the value of c.
x 1
1 
f ( x) 
on  , 2
x
2 
f is continuous on [ 1/2, 2 ], and differentiable on (1/2, 2).
1
f (2)  f  
2
1
2
2

3
3
2
3
2
 1
This should equal f ’(x)
at the point c. Now
find f ’(x).
x(1)  ( x  1)(1)
1
f '( x) 
 2
2
x
x
Determine if the mean value theorem applies, and if so
find the value of c.
x 1
1 
f ( x) 
on  , 2
x
2 
1
f (2)  f  
2
1
2
2

3
3
2
3
2
1
 2  1
x
x2  1
x 1
c 1
 1
x(1)  ( x  1)(1)
1
f '( x) 
 2
2
x
x
Application of
the Mean Value Theorem for Derivatives
You are driving on I 595 at 55 mph when you pass a
police car with radar. Five minutes later, 6 miles down the
road, you pass another police car with radar and you are
still going 55 mph. She pulls you over and gives you a
ticket for speeding citing the mean value theorem as
proof.
WHY ?
Application of the Mean Value Theorem for Derivatives
You are driving on I 94 at 55 mph when you pass a police car with radar. Five minutes later, 6
miles down the road you pass another police car with radar and you are still going 55mph. He
pulls you over and gives you a ticket for speeding citing the mean value theorem as proof.
Let t = 0 be the time you pass PC1. Let s = distance
traveled. Five minutes later is 5/60 hour = 1/12 hr. and
6 mi later, you pass PC2. There is some point in time
c where your average velocity is defined by
f b   f  a 
ba
s(1/12)  s(0)
6mi
Average Vel. =


(1/12  0)
1/12hr
72 mph
AP QUESTION
AP QUESTION