1
Rolle's Theorem (~ 𝟏𝟔𝟗𝟏) :
French mathematician Michel Rolle
(21 April 1652 – 8 November 1719)
If 𝑓(𝑥) is
① continuous on [a,b]
② differentiable on (a,b)
③ f(a) = f(b)
then, there ∃ c ϵ (a,b) such that f ’(c) = 0
(that means that there exists at least one)
Mean Value Theorem for Derivatives
French mathematician, Joseph-Louis Lagrange
(25 January 1736 – 10 April 1813)
If 𝑓(𝑥) is
① continuous on [a,b]
② differentiable on (a,b)
then, there ∃ c ϵ (a,b) such that 𝑓’(𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
Geometric Interpretation: Under the
given conditions, there is a point in
the open interval where the tangent
to the curve is the same as the slope
of the line joining the endpoints.
Application: Under the given conditions, there is a point in
the open interval where the instantaneous rate of change is
the same as the average rate of change on the interval (very
important). If the function is a position function, then there is
a point in the open interval where the instantaneous velocity
is the same as the average velocity on the interval
f'(c) 
f (b) f (a )
ba
is the average rate of change of the
function f(x) on the interval [a, b]
2
EX: Explain why Rolle’s Theorem does not apply for 𝑓(𝑥) = |𝑥 − 4| on [2, 5].
① 𝑓(𝑥) = |𝑥 − 4| is continuous on [2, 5].
② 𝑓(𝑥) = |𝑥 − 4| is not differentiable on (2, 5).
f(a) = f(b) does not need to be mentioned
𝑥
EX: Explain why Rolle’s Theorem does not apply for 𝑓(𝑥) = cot (2) on [𝜋, 3 𝜋].
𝑥
① 𝑓(𝑥) = cot (2) is not continuous on [𝜋, 3 𝜋]. .
② Therefore it is not differentiable on (𝜋, 3 𝜋).
f(a) = f(b) does not need to be checked
EX: Determine whether Rolle’s Theorem can be applied to 𝑓(𝑥) = 𝑥 2 − 3𝑥 on [0, 3].
If Rolle’s Theorem can be applied, find all values of c in the open interval (0, 3) such
that 𝑓 ′ (𝑐) = 0. If Rolle’s Theorem can not be applied, explain why.
① 𝑓(𝑥) = 𝑥 2 − 3𝑥 is continuous on [0, 3] (polynomial).
② 𝑓 ′ (𝑥) = 2𝑥 − 3 is defined for 𝑥 ∈ (0, 3).
→ 𝑓(𝑥) 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑜𝑛(0, 3).
③ 𝑓(0) = 𝑓(3) = 0
∴ 𝑤𝑒 𝑐𝑎𝑛 𝑎𝑝𝑝𝑙𝑦 𝑅𝑜𝑙𝑙𝑒 ′ 𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
2𝑥 − 3 = 0
→
𝑥=
3
𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑜𝑛 [0, 3].
2
EX: Determine whether Rolle’s Theorem can be applied to 𝑓(𝑥) = 𝑥 4 − 2𝑥 2 on [2, 3].
If Rolle’s Theorem can be applied, find all values of c in the open interval (2, 3)
such that 𝑓 ′ (𝑐) = 0. If Rolle’s Theorem can not be applied, explain why.
① 𝑓(𝑥) is continuous on [2, 3] (polynomial).
② 𝑓′(𝑥) = 4𝑥 3 − 4𝑥 is defined for 𝑥 ∈ (2, 3).
→ 𝑓(𝑥) 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑜𝑛 (2, 3).
③ 𝑓(2) = 8
𝑓(3) = 63 →
𝑓(2) ≠ 𝑓(3)
∴ 𝑤𝑒 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑎𝑝𝑝𝑙𝑦 𝑅𝑜𝑙𝑙𝑒 ′ 𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
Let us check for fun (Mrs. Radja’s kind of fun)
4𝑥 3 − 4𝑥 = 4𝑥(𝑥 2 − 1) = 4𝑥(𝑥 + 1)(𝑥 − 1) → 𝑥 = 0, ± 1
These values are not in (2, 3).
3
EX: Determine whether Rolle’s Theorem can be applied to 𝑓(𝑥) = (𝑥 − 3)(𝑥 + 1)2 on [-1, 3].
If Rolle’s Theorem can be applied, find all values of c in the open interval (-1, 3) such that
𝑓 ′ (𝑐) = 0. If Rolle’s Theorem can not be applied, explain why.
① 𝑓(𝑥) is continuous on [-1, 3] (polynomial).
② 𝑓 ′ (𝑥) = (1)(𝑥 + 1)2 + (𝑥 − 3)(2)(𝑥 + 1) = (𝑥 + 1)(3𝑥 − 5) is defined for 𝑥 ∈ (-1, 3).
→ 𝑓(𝑥) 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑜𝑛 (−1, 3).
③ 𝑓(−1) = 𝑓(3) = 0
∴ 𝑅𝑜𝑙𝑙𝑒 ′ 𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑
(𝑥 + 1)(3𝑥 − 5) = 0
Only 𝑥 =
5
3
→
𝑥 = −1, 𝑥 =
5
3
is on OPEN interval (-1, 3)
Function is not differentiable at x = -1, because definition of differentiability is:
Function is differentiable at a point if left derivative is equal to the right derivative at that point
EX: Determine whether Rolle’s Theorem can be applied to 𝑓(𝑥) =
𝑥 2 −1
on [-1, 1].
𝑥
If Rolle’s Theorem can be applied, find all values of c in the open interval (-1, 1) such that
𝑓 ′ (𝑐) = 0. If Rolle’s Theorem can not be applied, explain why.
① 𝑓(𝑥) is not continuous on [-1, 1].
② Therefore it is not differentiable on (-1, 1).
∴ 𝑤𝑒 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑎𝑝𝑝𝑙𝑦 𝑅𝑜𝑙𝑙𝑒 ′ 𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚
EX: Determine whether Rolle’s Theorem can be applied to 𝑓(𝑥) = 3 − |𝑥 − 3| on [0, 6].
If Rolle’s Theorem can be applied, find all values of c in the open interval (0, 6) such that
𝑓 ′ (𝑐) = 0. If Rolle’s Theorem can not be applied, explain why.
① 𝑓(𝑥) is continuous on [0, 6].
② 𝑓(𝑥) is not differentiable on (0, 6).
f(a) = f(b) does not need to be mentioned
∴ 𝑅𝑜𝑙𝑙𝑒 ′ 𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑏𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑
4
EX: Determine whether Rolle’s Theorem can be applied to 𝑓(𝑥) = −3𝑥√𝑥 + 1 on [-1, 0].
If Rolle’s Theorem can be applied, find all values of c in the open interval (0, 6) such that
𝑓 ′ (𝑐) = 0. If Rolle’s Theorem can not be applied, explain why.
① 𝑓(𝑥) is continuous on [-1, 0].
② 𝑓 ′ (𝑥) = −3√𝑥 + 1 − 3𝑥
1
2√𝑥+1
is defined for 𝑥 ∈ (-1, 0). (not defined for x = -1)
→ 𝑓(𝑥) 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑜𝑛 (−1, 0).
③ 𝑓(−1) = 𝑓(0) = 0
∴ 𝑅𝑜𝑙𝑙𝑒 ′ 𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑
−3√𝑥 + 1 = 3𝑥 2
→
1
√𝑥+1
− 6(𝑥 + 1) = 3𝑥
→
− 9𝑥 = 6
2
→ 𝑥 = − ∈ (−1, 0)
3
EX: Determine whether the MVT can be applied to f(x) = x3 – x on [0, 2]
● Since f is a polynomial, it is continuous and differentiable for all x.
● Therefore, by the MVT, there is a number c in (0,2) such that:
● f(2) = 6, f(0) = 0, and f ’(x) = 3x2 – 1.
6
2
● 3𝑐 2 − 1 =
→ 3𝑐 2 = 4 → 𝑐 = ±
2
√3
● However, c must lie in (0, 2), so 𝑐 =
2
√3
The tangent line at this value of c is parallel to the secant line OB.
If an object moves in a straight line with position function s = f(t), then the average velocity between t =
a and t = b is
f (b)  f ( a )
ba
and the velocity at t = c is f ’(c).
Thus, the MVT tells us that, at some time t = c between a and b, the instantaneous velocity f ’(c) is equal
to that average velocity.
● For instance, if a car traveled 180 km in 2 hours, the speedometer must
have read 90 km/h at least once.
In general, the MVT can be interpreted as saying that there is a number
at which the instantaneous rate of change is equal to the average rate of change over an interval.
EX: Determine whether MVT can be applied to 𝑓(𝑥) = 𝑥(𝑥 2 − 𝑥 − 2) on [-1, 1].
If yes, find all values of c on (-1, 1) such that 𝑓 ′ (𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
5
① 𝑓(𝑥) is continuous on [-1, 1]
② 𝑓 ′ (𝑥) = 3𝑥 2 − 2𝑥 − 2 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑜𝑛 (-1, 1) → 𝑓(𝑥) 𝑖𝑠 differentiable on (-1, 1)
3𝑐 2 − 2𝑐 − 2 =
−2−0
2
= −1 → 3𝑐 2 − 2𝑐 − 1 = 0
(3𝑐 + 1)(𝑐 − 1) = 0
→
𝑐=−
1
3
& 𝑥=1
𝑥+1
1
on [ , 2].
𝑥
2
𝑓(𝑏)−𝑓(𝑎)
= 𝑏−𝑎
EX: Determine whether MVT can be applied to 𝑓(𝑥) =
1
If yes, find all values of c on (2, 2) such that 𝑓 ′ (𝑐)
1
2
① 𝑓(𝑥) is continuous on [ , 2] (not continuous at 𝑥 = 0)
𝑑
1
1
1
1
② 𝑓 ′ (𝑥) = 𝑑𝑥 (1 + 𝑥) = − 𝑥 2 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑜𝑛 (2, 2) → 𝑓(𝑥) 𝑖𝑠 differentiable on (2, 2)
3
−3
1
− 2=2
= −1
3
𝑐
−2
1
→ 𝑐 = ±1 → 𝑐 = 1 ∈ ( , 2)
2
EX: Determine whether MVT can be applied to 𝑓(𝑥) = 2 sin 𝑥 + sin 2𝑥 on [0, 𝜋].
If yes, find all values of c on (0, 𝜋) such that 𝑓 ′ (𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
① 𝑓(𝑥) is continuous on [0, 𝜋]
② 𝑓 ′ (𝑥) = 2 cos 𝑥 + 2 cos 2𝑥 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑜𝑛 (0, 𝜋) → 𝑓(𝑥) 𝑖𝑠 differentiable on (0, 𝜋)
cos 2𝑥 = 𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑛2 𝑥
𝑓(𝜋) − 𝑓(0)
2 cos 𝑐 + 2 cos 2𝑐 =
=0
𝜋
= 2𝑐𝑜𝑠 2 𝑥 − 1
cos 𝑐 + 2𝑐𝑜𝑠 2 𝑥 − 1 = 0
(cos 𝑐 + 1)(2 cos 𝑐 − 1) = 0 → cos 𝑐 = −1 & cos 𝑐 =
→
𝑐=𝜋 & 𝑐=
𝜋
3
1
2