Download Section 3.2 – Rolle’s Theorem and the Mean Value Theorem

Section 3.2 – Rolle’s Theorem
and the Mean Value Theorem
Mean Value Theorem
The Mean Value Theorem can be interpreted geometrically as follows:
y  f  x
This is a continuous and
differentiable function.
 a, f  a  
 b, f  b  
f  b   f  a  Is the slope of the
line segment joining
ba
the points where x=a
and x=b.
 c, f  c  
MVT says at least one point like x=c
must exist. It does NOT find where it is.
f 'c
Is the slope of the line
tangent to the graph at
the point x=c. This is
equal to the slope of
the line segment.
Mean Value Theorem
Let f be a function that satisfies the following hypotheses:
1. f is continuous on the closed interval [a,b]
2. f is differentiable on the open interval (a,b)
Then there is a number c in (a,b) such that:
f 'c 
f b f  a 
ba
The derivative must equal the slope between the endpoints for at
least one value of x.
Example: If a car travels smoothly down a straight level road with
average velocity 60 mph, we would expect the speedometer
reading to be exactly 60 mph at least once during the trip.
There is at least one point where the instantaneous rate of change
equals the average rate of change.
White Board Challenge
Sketch a graph of the function with the
following characteristic:
The function is only defined on [2,8].
The function is differentiable on (2,8).
f(2) = f(8)
Does anyone have a graph that does not have
a horizontal tangent line?
Rolle’s Theorem
Let f be a function that satisfies the following three hypotheses:
There
must be at
1. f is continuous on the closed interval [a,b]
least one
2. f is differentiable on the open interval (a,b)
critical
3. f(a) = f(b)
number
Then there is a number c in (a,b) such that f '(c) = 0.
f(x)
between
the
endpoints.
At least one
critical point.
Start and end at
Example: If a car begins and ends at the same
place, y-value
then
the same
x
somewhere during
it must
reverse
direction.
a its journey,
b
c
Example 1
Show that the function f  x   x satisfies the hypotheses
of the MVT on the closed interval [9,25], and explain
what conclusions you can draw from it.
The function f is not differentiable at x≤0. BUT x≤0 is not in
the interval [9,25]. Thus, the function is continuous on
[9,25] and differentiable on (9,25). The hypotheses of the
MVT are satisfied.
Use the
Theorem:
f 'c 
f  25  f  9 
259

25  9
16
 162  18
Therefore the Mean Value Theorem guarantees at least
one x–value in (9,25) at which the instantaneous rate of
change of f is equal to 1/8.
Example 2
3
2
f
x

x

x
Show that the function  
satisfies the
hypotheses of the MVT on the closed interval [1,2], and
find a number c between 1 and 2 so that:
f  2   f 1
f 'c 
2 1
Because f is a polynomial function, it
is differentiable and also continuous
on the entire interval [1,2]. Thus, the
hypotheses of the MVT are satisfied.
Find the derivative.
f '  x   dxd x3  x 2

f '  x   3x 2  2 x

c
2 
Find the value of c.
f  2   f 1
f '  c   21
 2  2 1 1 
3
2
3
3c  2c 
1
2
3c  2c  10
3c 2  2c  10  0
 1.523
 2 2  4 3 10 
2 124

6
2 3
 2.189
2
2
Example 3
Discuss why Rolle’s Theorem can not be applied to the
functions below.
f  x  x  2
on 0, 4
Not differentiable at
x=2
f  x   tan x on 0, 2 
Not continuous
2007 AB Free Response 3
We already learned that
this implies continuity.
Notice how every part of
the MVT is discussed
(Continuity AND
Differentiability).
The functions f and g are differentiable for all real numbers, and g is
strictly increasing. The table above gives values of the functions and
their first derivatives at selected values of x. The function h is given by
h(x) = f(g(x)) – 6.
Explain why there must be a value of c for 1< c < 3 such that h'(c) = -5.
h  3  h 1
31
f g  3  f g 1
  2   
f  4 f  2
2

19
2
 5
Since h is continuous and
differentiable, by the MVT, there
exists a value c, 1 < c < 3, such
that h' (c) = -5.