Mean Value Theorem - Metropolitan Community College

Mean Value Theorem
Joseph Lee
Metropolitan Community College
Joseph Lee
Mean Value Theorem
Rolle’s Theorem
Let f be a continuous function defined on a closed interval [a, b]
that is differentiable on the open interval (a, b). If f (a) = f (b)
then there exists some real number c in the open interval (a, b)
such that f 0 (c) = 0.
Joseph Lee
Mean Value Theorem
Example 1.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = 4x − x 2 ,
[0, 4]
Joseph Lee
Mean Value Theorem
Example 1.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = 4x − x 2 ,
[0, 4]
Solution. Since f is differentiable everywhere and
f (0) = f (4) = 0, Rolle’s Theorem applies.
f 0 (x) = 4 − 2x = 0
x= 2
Thus, for c = 2, f 0 (c) = 0.
Joseph Lee
Mean Value Theorem
Example 2.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = x 3 + 3x 2 − x − 1,
[−3, 1]
Joseph Lee
Mean Value Theorem
Example 2.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = x 3 + 3x 2 − x − 1,
[−3, 1]
Solution. Since f is differentiable everywhere and
f (−3) = f (1) = 2, Rolle’s Theorem applies.
f 0 (x) = 3x 2 + 6x − 1 = 0
√
−3 ± 2 3
x=
3
√
−3 ± 2 3 0
Thus, for c =
, f (c) = 0.
3
Joseph Lee
Mean Value Theorem
Example 3.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = sin x,
[0, π]
Joseph Lee
Mean Value Theorem
Example 3.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = sin x,
[0, π]
Solution. Since f is differentiable everywhere and
f (0) = f (π) = 0, Rolle’s Theorem applies.
f 0 (x) = cos x = 0
x=
Thus, for c =
π
2
π 0
, f (c) = 0.
2
Joseph Lee
Mean Value Theorem
Example 4.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = tan x,
[0, π]
Joseph Lee
Mean Value Theorem
Example 4.
Determine if Rolle’s Theorem is applicable on the given closed
interval. If Rolle’s Theorem is appicable, find all values c such that
f 0 (c) = 0.
f (x) = tan x,
[0, π]
Solution. Rolle’s Theorem does not apply. f is not differentiable
π
at .
2
Joseph Lee
Mean Value Theorem
Mean Value Theorem
Let f be a continuous function defined on a closed interval [a, b]
that is differentiable on the open interval (a, b). Then there exists
some real number c in the open interval (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
Joseph Lee
Mean Value Theorem
Example 5.
Determine if the Mean Value Theorem is applicable on [a, b]. If
Rolle’s Theorem is appicable, find all values c such that
f (b) − f (a)
.
f 0 (c) =
b−a
f (x) = 4x − x 2 ,
Joseph Lee
[0, 5]
Mean Value Theorem
Example 5.
Determine if the Mean Value Theorem is applicable on [a, b]. If
Rolle’s Theorem is appicable, find all values c such that
f (b) − f (a)
.
f 0 (c) =
b−a
f (x) = 4x − x 2 ,
[0, 5]
Solution. Since f is differentiable everywhere, the Mean Value
Theorem applies.
f 0 (x) = 4 − 2x =
f (5) − f (0)
5−0
4 − 2x = −1
5
2
5
f (b) − f (a)
.
Thus, for c = , f 0 (c) =
2
b−a
x=
Joseph Lee
Mean Value Theorem
Example 6.
Determine if the Mean Value Theorem is applicable on [a, b]. If
Rolle’s Theorem is appicable, find all values c such that
f (b) − f (a)
.
f 0 (c) =
b−a
f (x) = x 3 ,
Joseph Lee
[−1, 1]
Mean Value Theorem
Example 6.
Determine if the Mean Value Theorem is applicable on [a, b]. If
Rolle’s Theorem is appicable, find all values c such that
f (b) − f (a)
.
f 0 (c) =
b−a
f (x) = x 3 ,
[−1, 1]
Solution. Since f is differentiable everywhere, the Mean Value
Theorem applies.
f 0 (x) = 3x 2 =
f (1) − f (−1)
1 − (−1)
3x 2 = 1
1
x = ±√
3
1
f (b) − f (a)
Thus, for c = ± √ , f 0 (c) =
.
b−a
3
Joseph Lee
Mean Value Theorem