The Mean Value Theorem
Math 131
Spring 2010
Max-Min Theorem
If f'(x) is defined on an open interval (a, b) and if
f(x) has a relative extrema at a point c on the
interval, then f'(c) = 0
Note that f'(c) = 0 is only a necessary and not a
sufficient condition; the existence of a zero
derivative does not guarantee a relative extrema
Proof (outline)
f c  h  f c 
If f   c   lim
exists then both the
h0
h
f c  h  f c
left hand limit f   c   hlim
and the
0
h
f c  h  f c
right hand limit exist f   c   hlim
0
h
Assume that f  c  is a local (relative) maximum
f c  h  f c
In the left hand limit f   c   hlim
0
h
f  c  h  is always less than or equal to f  c 
since f  c  is a relative maximum. Therefore
the left hand derivative (?)
f c  h  f c
f   c   lim
0
h0
h
is always less than or equal to 0.
Using a similar argument one can show that the
right hand derivative is always greater than or
equal to 0
f c  h  f c 
lim
0
h0
h
Therefore since the left and right hand limits
exist and equal each other, we conclude that
the derivative equals 0 – their “common value”.
f   c   lim
h0
f c  h  f c 
0
h
Rolle’s Theorem
If f’(x) is continuous on the closed interval [a, b]
and differentiable on the open interval (a, b)
and if f(a) = 0 = f(b) then somewhere on the
open interval (a, b) there is a point c such that
f‘(c) = 0.
This result follows from the Max-Min theorem.
Proof of Rolle’s Theorem
Either f(x) is constant on [a, b] which means its
derivative f’(x) = 0 everywhere
Or starting from f(a) = 0, f(x) increases on the
interval (a, b) only to “turn around” (relative
maximum) to return to f(b) = 0
Or starting from f(a) = 0, f(x) decreases on the
interval (a, b) only to “turn around” (relative
minimum) to return to f(b) = 0.
In the last two cases the Max-Min theorem says
f ‘(c) = 0 at the relative maximum/minimum.
The only real application of Rolle’s Theorem is
using it in the proof of the Mean Value
Theorem
Mean Value Theorem
If f(x) is continuous on the closed interval [a, b]
and differentiable on the open interval (a, b)
then there is a point c on the open interval
(a, b) such that
f b   f  a 
f c  
ba
That is, somewhere there is a derivative equal
to the average rate of change over the interval
In the next diagram the average rate of change
of a function f(x) over the interval [a, b] is seen
as the slope of the blue secant line.
Observe that we can move the blue secant line
“in parallel” so that it touches the curve at a
single point c – becoming a (red) tangent line
whose slope is f’(c).
Thus the Mean Value Theorem
mtan = f’(c)
f(x)
a
c
b
Proof:
f (b)  f (a)
( x  a )   f ( x)  f (a ) 
Define: g ( x) 
ba
Observe: g (a)  0  g (b)
Therefore by Rolle’s Theorem there is a point c
on the interval (a, b) such that
f (b)  f (a )
g (c) 
 f (c)  0
ba
f (b)  f (a)
Solving for f’(c) yields f (c) 
ba
QED!
Example: f ( x)  4  x 2 is differentiable on (-1, 3).
f (3)  f (1)
Find the point c where f '(c) 
and
3  (1)
compute the equation of the tangent line.
f (3)  f (1) 4  3  (4  (1) ) 5  (3)
f c  


 2
3  (1)
4
4
2
2
where f ( x)  2 x . Solving 2c  2 yields c  1
Since f (1)  3 the equation of the tangent line is
y  2( x  1)  3  2 x  5
The MVT: Urban Legends
A driver enters the State Turnpike at 2:00 PM
and at 4:00 PM exits at a point 140 miles away.
His average speed is 70 mph. The posted speed
limit on the turnpike is 65 mph. Therefore by the
Mean Value Theorem, somewhere his
speedometer (instantaneous rate of change)
equaled his average rate change – 70 mph.
The driver was immediately arrested - for
speeding!
The MVT: Urban Legends
Although there have been rumors that some
states have sought to apply the Mean Value
Theorem to catch speeders, there is no hard
evidence that this was ever done (successfully).
See: Snopes.com: E-ZPass Speeding Tickets
http://www.snopes.com/autos/law/ezpass.asp
Another Application of the MVT
Proving if f’(x) = 0 on an interval then f(x) is a constant
function
Suppose f’(x) = 0 on some interval I but f(x) is
not a constant function.
Therefore there are two points a and b on
interval I such that f(a) ≠ f(b)
Therefore the average rate of change on the
interval [a,b] is
f (b)  f (a)
 0 <- not zero!
ba
Another Application of the MVT
Proving if f’(x) = 0 on an interval then f(x) is a constant
function (cont.)
By the Mean Value Theorem there is a point c on
the interval [a,b] such that
f (b)  f (a)
f c  
0
ba
But wait a minute! f’(c) = 0 so something is
wrong! Where did we go wrong? We assumed
that f(x) is not a constant function
Therefore f(x) is constant on the interval I. QED!