4.1 Antiderivateves and Indefinite Integrals

3.2 Rolles &
Mean Value
Theorem
Rolle’s Theorem
Let f be continuous on the closed
interval [a,b] and differentiable on
the open interval (a,b). If
𝑓 π‘Ž = 𝑓(𝑏)
Then there is at least one number c
in (a,b) such that f’(c)=0
What does Rolle’s Thrm do?

Rolle’s theorem states some x value exists
(x=c) so that the tangent line at that specific
x value is a horizontal tangent (f’(c)=0)
Horizontal Tangent Line
ie: f’(c)=0
f(a)=f(b)
a
c
b
Notes about Rolle’s Thrm

It is an EXISTENCE Theorem, it simply states that some c
has to exist. It does NOT tell us exactly where that
value is located.

In order to find the location x=c, we would take f’(x)=0
and find critical numbers like in section 3.1 (Extrema on
a closed Interval)
Example 1 of Rolle’s Thrm

Determine whether Rolle’s thrm can be applied. If it
can be applied, find all values of c such that f’(c)=0

Ex: 𝑓 π‘₯ = π‘₯ 2 βˆ’ 5π‘₯ + 4 , [1,4]

Since f is a polynomial, it is continuous on [1,4] and
differentiable (1,4).

𝑓 1 = 0 = 𝑓(4)

Therefore, Rolle’s Theorem can be applied and states
there must be some x=c on [1,4] such that f’(c)=0.

Lets find those x values!
Example 1 Continued

𝑓 β€² π‘₯ = 2π‘₯ βˆ’ 5

0 = 2π‘₯ βˆ’ 5

π‘₯=2

Therefore by Rolle’s Thrm, 𝑓 β€²
5
5
2
=0
Example 2 of Rolle’s Thrm

Determine whether Rolle’s thrm can be applied. If it
can be applied, find all values of c such that f’(c)=0

Ex: 𝑓 π‘₯ = π‘₯ + 1 π‘œπ‘› [βˆ’2,3]

f is continuous on [-2,3]

f is not differentiable on (-2,3)

ROLLE’s cannot be used!
Example 3 of Rolle’s Thrm

Determine whether Rolle’s thrm can be applied. If it
can be applied, find all values of c such that f’(c)=0

Ex: 𝑓 π‘₯ = cos π‘₯

f is continuous on [0,2πœ‹]

f is differentiable on (0,2πœ‹)

cos 0 = 1 = cos 2πœ‹

ROLLE’S APPLIES!
[0,2πœ‹]
Example 3 Continued

𝑓 β€² π‘₯ = βˆ’sin(π‘₯)

βˆ’ sin π‘₯ = 0

π‘₯ = 0, πœ‹

Therefore 𝑓 β€² 0 = 0 π‘Žπ‘›π‘‘ 𝑓 β€² πœ‹ = 0
Mean Value Theorem

If f is continuous on [a,b] and differentiable on the open
interval (a,b), then there exists a number c in (a,b) s.t.

𝑓′ 𝑐 =
𝑓 𝑏 βˆ’π‘“ π‘Ž
π‘βˆ’π‘Ž
𝑓′ 𝑐
f(b)
𝑓 𝑏 βˆ’π‘“ π‘Ž
π‘βˆ’π‘Ž
f(a)
a
b
MVT Example
pg 177 #37 parts a,b,c,d