Rolle’s Theorem/Mean-Value
Theorem
Objective: Use and interpret the
Mean-Value Theorem
Rolle’s Theorem
• Rolle’s Theorem is a special case of the Mean-Value
Theorem.
• Theorem 5.7.1 (Rolle’s Theorem)
• Let f be continuous on the closed interval [a, b] and
differentiable on the open interval (a, b). If f(a) = 0
and f(b) = 0, then there is at least one point c in the
interval (a, b) such that f /(x) = 0
Example 1
• Find the two intercepts of the function f ( x)  x 2  5x  4
and confirm that f /(c) = 0 at some point between
those intercepts.
Example 1
• Find the two intercepts of the function f ( x)  x 2  5x  4
and confirm that f /(c) = 0 at some point between
those intercepts.
• If we factor the expression, we get (x – 1)(x – 4), so
the zeros are x = 1 and x = 4. To confirm the
hypothesis of Rolle’s Theorem, f(1) = f(4) = 0. We
know that there must a point between these two
values where f /(c) = 0.
Example 1
• Find the two intercepts of the function f ( x)  x 2  5x  4
and confirm that f /(c) = 0 at some point between
those intercepts.
• If we factor the expression, we get (x – 1)(x – 4), so
the zeros are x = 1 and x = 4. To confirm the
hypothesis of Rolle’s Theorem, f(1) = f(4) = 0. We
know that there must a point between these two
values where f /(c) = 0.
f / ( x)  2 x  5 c  5 / 2
f (5 / 2)  0
/
Example 2
• The differentiability requirement in Rolle’s Theorem is
critical. If f fails to be differentiable at even one place
in the interval (a, b), then the conclusion of the
theorem may not hold. For example, the function
f ( x) | x | 1 graphed below has f(-1) = f(1) = 0, yet
there is no horizontal tangent to the graph of f over
the interval (-1, 1).
Hypothesis of Rolle’s Theorem
• The homework will ask you first to confirm the
hypothesis of Rolle’s Theorem. They want you to say
that:
• the function is continuous and differentiable over the
interval (a, b)
• f(a) = f(b) = 0.
Example3
• If f satisfies the conditions of Rolle’s Theorem on [a, b], then
the theorem guarantees the existence of at least one point c in
(a, b) at which f /(c) = 0. There may, however, be more than
one such c. For example, the function f(x) = sinx is continuous
and differentiable everywhere, so the hypothesis of Rolle’s
Theorem are satisfied on the interval [0, 2p] whose endpoints
are roots of f [f(0) = f (2p) = 0]. We can see that there are two
points that satisfy the theorem, x = p/2 and x = 3p/2.
Mean-Value Theorem
• Rolle’s Theorem is a special case of a more general
result called the Mean-Value Theorem.
Geometrically, this theorem states that between any
two points A(a, f(a)) and B(b, f(b)) on the graph of a
differentiable function f, there is at least one place
where the tangent line to the graph is parallel to the
secant line joining A and B.
Mean-Value Theorem
• Note that the slope of the secant line joining A and B
is f (b)  f (a) .
ba
/
f
• The slope of the tangent line at c is (c) .
• This leads us to The Mean-Value Theorem:
• Theorem 5.7.2
• Let f be continuous on the closed interval [a, b] and
differentiable on the open interval (a, b). Then there
is at least one point c in (a, b) such that
f (b)  f (a )
f (c ) 
ba
/
Example 4
f ( x) 
1 3
x 1
4
• Show that the function
satisfies the
hypothesis of the Mean-Value Theorem over the
interval [0, 2], and find all values of c in the interval
(0, 2) at which the tangent line to the graph of f is
parallel to the secant line joining the points (0, f(0))
and (2, f(2)).
Example 4
f ( x) 
1 3
x 1
4
• Show that the function
satisfies the
hypothesis of the Mean-Value Theorem over the
interval [0, 2], and find all values of c in the interval
(0, 2) at which the tangent line to the graph of f is
parallel to the secant line joining the points (0, f(0))
and (2, f(2)).
• The function is continuous and differentiable
everywhere since it is a polynomial, so the hypothesis
is satisfied.
Example 4
f ( x) 
1 3
x 1
4
• Show that the function
satisfies the
hypothesis of the Mean-Value Theorem over the
interval [0, 2], and find all values of c in the interval
(0, 2) at which the tangent line to the graph of f is
parallel to the secant line joining the points (0, f(0))
and (2, f(2)).
2
• f(0) = 1
3 1
3 2
/
x
f (c ) 
1
x 1
3
20
4
• f(2) = 3
• The only answer in the interval is c  2
3
Example 4
f ( x) 
1 3
x 1
4
• Show that the function
satisfies the
hypothesis of the Mean-Value Theorem over the
interval [0, 2], and find all values of c in the interval
(0, 2) at which the tangent line to the graph of f is
parallel to the secant line joining the points (0, f(0))
and (2, f(2)).
2
c
3
Velocity
• There is a nice interpretation of the Mean-Value Theorem in
the situation where x = f(t) is the position vs. time curve for a
car moving along a straight road. In this case, the right side of
the equation is the average velocity of the car over the
interval from a to b and the left side is the instantaneous
velocity at time t = c. Thus, the Mean-Value Theorem implies
that at least once during the time interval the instantaneous
velocity must equal the average velocity.
f (b)  f (a )
f (c ) 
ba
/
Example 5
• You are driving on a straight highway on which the
speed limit is 55 mi/h. At 8:05 AM a police car clocks
your velocity at 50 mi/h and at 8:10 AM a second
police car posted 5 mi down the road clocks your
velocity at 55 mi/h. Explain why the police have a
right to charge you with a speeding violation.
Example 5
• You are driving on a straight highway on which the
speed limit is 55 mi/h. At 8:05 AM a police car clocks
your velocity at 50 mi/h and at 8:10 AM a second
police car posted 5 mi down the road clocks your
velocity at 55 mi/h. Explain why the police have a
right to charge you with a speeding violation.
• You traveled 5 miles in 5 minutes, or a mile per
minute, or 60 mi/h. The Mean-Value Theorem
guarantees the police that your instantaneous
velocity was 60 mi/h at some point.
Homework
• Pages 334-335
• 1-11 odd
• 15, 19