Section 4.2 The Mean Value Theorem
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Rolle’s Theorem
Theorem (Rolle’s Theorem)
Let f (x) be a function that satisfies:
1. f is continuous on the closed interval [a, b].
2. f is differentiable on the open interval (a, b).
3. f (a) = f (b).
Then there is a number c in (a, b) such that f 0 (c) = 0.
y
y
f'(c1)=0
y=f(x)
f'(c3)=0
b x
a
1
f'(c2)=0
a
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b
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x
y=|x|
-1
1
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Proof of Rolle’s Theorem
Let f (a) = f (b) = k. There are three cases:
If f takes on a local maximum value at a number c in (a, b). Then Fermat’s
Theorem guarantees f 0 (c) = 0.
If f takes on a local minimum value at a number c in (a, b). Then Fermat’s
Theorem guarantees f 0 (c) = 0.
Otherwise there is not a local extreme value in (a, b). According to the
Extreme Value Theorem, the function f must take on its absolute extreme
values at the endpoints. Thus both absolute maximum and minimum value
are the same value k. This implies that f (x) = k is a constant. We have
f 0 (x) = 0 for each x in (a, b).
y
f'(c)=0
y
y=f(x)
y=f(x)
k
k
f'(c)=0
x
x
a
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a
b
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b
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Example
Example
Show that the only solution to the equation x = arctan x is x = 0.
Solution Let f (x) = x − arctan x. If there were another solution b to the original
equation, we would have
f (0) = f (b) = 0.
By Rolle’s Theorem, there would be a number c between 0 and b, such that
f 0 (c) = 0. But the derivative
f 0 (x) =
1
x2
d
(x − arctan x) = 1 −
=
>0
dx
1 + x2
1 + x2
for any x 6= 0 . This is a contradiction.
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MATH 1ZA3
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The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be a function that satisfies
1. f is continuous on the closed interval [a, b].
2. f is differentiable on the open interval (a, b).
Then there is a number c in (a, b) such that
f 0 (c) =
f (b) − f (a)
b−a
= the average rate of change of f
or equivalently, f (b) − f (a) = f 0 (c)(b − a).
Remark 1 If f (x) = kx + q is a linear function, then we have
f 0 (x) = k =
f (b) − f (a)
b−a
for any x ∈ (a, b). Thus the average rate of change is the only choice that
guarantees the existence of c in the theorem.
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Proof of the Mean Value Theorem
Step 1 Let us define a linear function
L(x) = f (a) +
f (b) − f (a)
· (x − a).
b−a
It satisfies L(a) = f (a) and
L(b) = f (a) +
f (b) − f (a)
· (b − a) = f (a) + [f (b) − f (a)] = f (b).
b−a
Step 2 Consider the function g (x) = f (x) − L(x). The function g (x) is still
continuous on the closed interval [a, b] and differentiable on the open interval
(a, b). In addition, we have g (a) = g (b) = 0. According to Rolle’s Theorem,
there exists a number c in (a, b) such that g 0 (c) = 0. By the fact
g 0 (c) = f 0 (c) − L0 (c), we have
g 0 (c) = 0 =⇒ f 0 (c) = L0 (c) =
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f (b) − f (a)
.
b−a
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Illustration of the Mean Value Theorem
y
f'(c)=
f(b)-f(a)
b-a
y=f(x)
(f,f(b))
(c,f(c))
g(x) = f(x)-L(x)
y=L(x)
slope =
f(b)-f(a)
b-a
(a,f(a))
x
a
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b
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Example
Example
Prove the inequality x > sin x holds for any 0 < x < π/2.
Proof Let f (x) = x − sin x. This function is continuous and differentiable
everywhere with
f 0 (x) = 1 − cos x.
Given any number b ∈ (0, π/2), we can apply the Mean Value Theorem on the
function f in the interval [0, b] and conclude that there exists a number c in
(0, b), such that
f (b) − f (0) = f 0 (c)(b − 0) = (1 − cos c)b > 0
This implies f (b) > f (0) = 0 ⇒ b > sin b. Therefore we obtain x > sin x for any
x ∈ (0, π/2).
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An Application of the Mean Value Theorem
Theorem
(a) If f 0 (x) = 0 for all x in an interval (a, b), then f is constant on (a, b).
(b) If f 0 (x) = g 0 (x) for all x in an interval (a, b), then f − g is constant on (a, b);
that is, f (x) = g (x) + C where C is a constant.
Proof (a) Let x1 < x2 be two different numbers in the interval (a, b). According
to the Mean Value Theorem, there exists a number c in (x1 , x2 ), such that
f (x2 ) − f (x1 ) = f 0 (c) · (x2 − x1 ) = 0.
Therefore we have f (x2 ) = f (x1 ). This argument works for any two numbers
x1 < x2 in the interval (a, b). Thus the function f is constant in the interval (a, b).
(b) If we define h(x) = f (x) − g (x), then h0 (x) = f 0 (x) − g 0 (x) = 0 for all x in
(a, b). By the part (a), we conclude that h, namely f − g , is constant in (a, b).
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Examples
Example
Find all functions f (x) that satisfies f 0 (x) = cos x.
d
(sin x). By the theorem, we obtain
dx
f (x) = sin x + C . Here C is a constant.
Solution We have f 0 (x) = cos x =
Example
Let f (x) be a differentiable function that satisfies |f 0 (x)| ≤ m for all x. Prove that
the inequality |f (x2 ) − f (x1 )| ≤ m|x2 − x1 | holds for any two numbers x1 < x2 .
Proof Given x1 < x2 , by the Mean Value Theorem, there exists a number c in
(x1 , x2 ), such that
f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 )
Therefore we have
|f (x2 ) − f (x1 )| = |f 0 (c)| · |x2 − x1 | ≤ m|x2 − x1 |.
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The Second Mean Value Theorem (Optional)
Theorem
Let f (x), g (x) be two functions that satisfy:
1. f , g are continuous on the closed interval [a, b].
2. f , g are differentiable on the open interval (a, b).
3. g 0 (x) 6= 0 for x ∈ (a, b).
Then there is a number c in (a, b) such that
f 0 (c)
f (b) − f (a)
=
g 0 (c)
g (b) − g (a)
f (b) − f (a)
g (x). It is still continuous
g (b) − g (a)
in the closed interval [a, b] and differentiable in the open interval (a, b). In
addition, we have h(b) = h(a). By Rolle’s Theorem, there exists a number c in
(a, b), such that
Proof Consider the function h(x) = f (x) −
h0 (c) = 0 =⇒ f 0 (c) −
Ruipeng Shen
f (b) − f (a) 0
f 0 (c)
f (b) − f (a)
g (c) = 0 =⇒ 0
=
.
g (b) − g (a)
g (c)
g (b) − g (a)
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