Mean Value Theorem
The Mean Value Theorem connects the average rate of change (slope of the secant
between two points [a and b]) with the instantaneous rate of change (slope of
tangent at some point c).
The theorem says that somewhere between a and b on a differentiable curve, there is
at least one tangent line parallel to the slope of the secant between points a and b.
Definition: Mean Value Theorem
If
y  f  x  is continuous at every point of the closed interval  a, b and
differentiable at every point of its interior
in
 a, b 
at which
f c 
 a, b  , then there is at least one point c
f b  f  a 
ba
Physically, the Mean Value Theorem says that the instantaneous change at some
interior point must equal the average change over the entire interval.
Example
a) Show that
f  x   x 2  2 x  1 satisfies the Mean value Theorem on the interval
0, 1
b) Find the value c in (0, 1) that satisfies the equation.
Solution
a) Since f is continuous on
0, 1
and differentiable (0, 1). Since
f  0   1 and
f 1  2 , the Mean Value Theorem guarantees a point c in the interval (0, 1)
f c 
b)
2c  2 
f 1  f  0 
1 0
2   1
2c  1
1
c
2
1
Example
The function
 x,
f  x  
0,
0  x 1
x 1
Is zero at x=0 and at x=1. Its derivative is equal to 1 at every point between 0 and 1,
so f  is never zero between 0 and 1, and the graph of f has no tangent parallel to the
chord from (0, 0) to (1, 0). Explain why this does not violate the Mean Value
Theorem.
Solution
Because the function is not continuous on
0, 1 ; the function does not satisfy the
hypothesis of the Mean Value Theorem and need not satisfy the conclusions.
Example
The geometric mean of any two positive number a and b is
ab . Show that if
2
on any interval [a, b], then the value of c in the conclusion of the Mean
x
Value theorem is c  ab
f  x 
Solution
2 2
f b  f  a  b  a
2


ba
ba
ab
f c  

2
c2
2
2

2
c
ab
c 2  ab
c  ab
Positive since ‘a’ and ‘b’ are both
positive, and c is between ‘a’ and
‘b’ so it also must be positive.
Example
Suppose f is differentiable in (a, b) and
f  x1   f  x2   M x1  x2 for all x1 , x2   a, b 
a) Prove that
b) Prove
sin  a   a , a
c) Using a), prove
sin  x1   sin  x2   x1  x2
Solution
a) From Mean Value theorem
f  x  M
f  x1   f  x2 
M
x1  x2
f  x1   f  x2 
x1  x2
M
f  x1   f  x2   M x1  x2
b)
sin  a   sin  a   sin  0 
sin  a   sin  0 
 f   c  , c   a,0 
a0
sin  a   sin  0 
 f c
a0
sin  a 
 cos  c 
a
sin  a 
1
a
sin  a   a
c)
f   x   M for all x   a, b 
sin   x   M
cos  x   M
1 M
f  x1   f  x2   M x1  x2
sin  x1   sin  x2   M x1  x2
M x1  x2  x1  x2
Definition: Increasing and Decreasing Functions
Let f be continuous on


 a, b 
and differentiable on
 a, b 
f   0 at each point of (a, b), then f increases on [a, b]
If f   0 at each point of (a, b), then f decreases on [a, b]
If
Corollary: Functions with the same derivative differ by a constant
Example
Prove that
x3  4 x  2  0 has exactly one solution.
Solution:
Since the derivative 3x  4 is always positive, then the function is always increasing.
Therefore the function cannot have two solutions as it does not have a change of
direction (does not decrease).
2
Definition: Antiderivative
A function F(x) is an antiderivative of a function f(x) if
F   x   f  x  for all x in
the domain of f. The process of finding the Antiderivative is anti-differentiation.
Example
f  x   x 4  x 4
Find the Antiderivative of
Solution:
1 5 1 3
x 
x C
5
3
1
1
 x5  x 3  C
5
3
F  x 
Example
f  x   2cos  x   3
Find the Antiderivative of
Solution:
F  x   2sin  x   3x  C
Example
f  x   10e5 x  2e3 x
Find the Antiderivative of
Solution:
10 5 x 2 3 x
e  e C
5
3
2
 2e5 x  e3 x  C
3
F  x 
Example
Find the Antiderivative of
f  x   3sin
Solution:
F  x  
3
cos
3
 3x  
1
e
3
3x
C
 3x   e
3x
Example
On the moon, the acceleration due to gravity is
1.6
m
sec 2
a) If a rock is dropped into a crevasse, how fast will it be going just before it hits
bottom 20 seconds later.
b) How far below the point of release is the bottom of the crevasse?
c) If instead of being released from rest, the rock is thrown into the crevasse from the
same point with a downward velocity of
3
m
, when will it hit the bottom and how
sec
fast will it be going when it does?
Solution
a) Since acceleration the derivative of the velocity, then
v  t   1.6
Now finding the Antiderivative will find the velocity function v(t).
v  t   1.6t  C
We need to find the value of c.
We know that when t=0, v(t)=0
v  0   1.6  0   C
0  0C
C 0
Therefore
v  t   1.6t and v  20   1.6  30 
 32
Thus the rock will be travelling at 32 m/sec.
b) Let
s  t  represent the position of the rock at any time t.
Since
Now
Thus
s  t   v  t 
 1.6t
s  t   0.8t 2  C
s  0   0 , therefore C=0, so s  t   0.8t 2
s  20   0.8  20 
 320 m
2
c) As in above
v  t   1.6t  C
v  0   1.6  0   C
3C
Therefore
v  t   1.6t  3
s  t   1.6t  3
s  t   0.8t 2  3t  C
s (0)  0.8  0   4  0   C
0C
 s  t   0.8t 2  3t
2
Now we know that the depth is 320 metres
Therefore 320  0.8t
 3t
0.8t  3t  320  0
t  21.963, t  18.213
2
2
The velocity is
v 18.213  32.141
This tells us that the rock will hit the bottom after about 18.213 seconds and will hit
with velocity of about 32.141 m/sec
Example
A ball is dropped from a height 45 m above the ground. At the same instant another
ball is thrown downward with a velocity of 20 m/s from a height of 60m above the
ground. Which ball hits the ground first given that the acceleration due to gravity is 10
m/s2.
Solution:
In this question, we choose down to be negative. Therefore my acceleration will be
negative, and the tossed ball with have a negative velocity.
Let’s find the velocity and position functions of the dropped ball.
a  t   10
v  t   10t  C
0  10  0   C
C 0
 v  t   10t
s  t   5t 2  C
0  5  0   C
C 0
 s  t   5t 2
2
Now let’s find the time when the position is minus 45
45  5t 2
9  t2
t  3 [omit -3]
Let’s find the velocity and position functions of the thrown ball.
a  t   10
v  t   10t  C
20  10  0   C
C  20
 v  t   10t  20
s  t   5t 2  20t  C
0  5  0   20  0  C
C 0
 s  t   5t 2  20t
2
Now let’s find the time when the position is minus 45
60  5t 2  20t
5t 2  20t  60  0
t 2  4t  12  0
 t  6  t  2   0
t  2 [omit -6]
Time difference is 3-2=1
Therefore the thrown ball will hit the 1 second before the dropped ball.