Prem Gurung,
Pokhara University
Mean value theorem
Prem Gurung
Assistant Professor
School of Engineering, Faculty of Science and Technology
Pokhara University, Kaski Nepal
Rolle's Theorem:
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Pokhara University
Suppose that a function 𝑓(𝑥) is continuous on the closed interval [𝑎, 𝑏] and
differentiable on the open interval (𝑎, 𝑏). Then if 𝑓(𝑎) = 𝑓(𝑏), then there exists
at least one point 𝑐 in the open interval (𝑎, 𝑏) for which 𝑓′(𝑐) = 0.
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Proof:
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Since 𝑓(𝑥) is continuous on closed interval 𝑎, 𝑏 , then by extreme value
theorem, 𝑓(𝑥) has both a minimum and maximum on the interval [𝑎, 𝑏]. Let 𝑀
be the maximum and 𝑚 be the minimum value on the interval [𝑎, 𝑏].
Case I:
If 𝑀 = 𝑚, i.e. 𝑓 𝑥 is constant in [a, b]. Then obviously 𝑓′(𝑥) = 0 at every point
in the interval and 𝑐 can be taken any real number in (a, b). That is theorem is
true in this case.
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Case II:
If 𝑀 ≠ 𝑚 then at least one of them is different from 𝑓 𝑎 or 𝑓 𝑏 .
let f(c) is a maximum value of f(x) in (a, b). (i.e. f a ≠ 𝑀) we wish to show
that 𝑓′ 𝑐 = 0 for some 𝑐 in (a, b).
Suppose that 𝑓 𝑐 = M, then
𝑓 𝑐 − ℎ ≤ 𝑓(𝑐) and 𝑓 𝑐 + ℎ ≤ 𝑓 𝑐
for all h > 0
So, [𝑓 𝑐 − ℎ − 𝑓(𝑐) ≤ 0)] and
[𝑓 𝑐 + ℎ − 𝑓(𝑐) ≤ 0)] for all h > 0
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𝑓 𝑐+ℎ −𝑓(𝑐)
ℎ
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=
−𝑣𝑒
+𝑣𝑒
≤0
and
𝑓 𝑐−ℎ −𝑓(𝑐)
−ℎ
≤0
and
𝑓 𝑐−ℎ −𝑓(𝑐)
−ℎ
ℎ→𝑜
=
−𝑣𝑒
−𝑣𝑒
≥ 0 for all h > 0
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𝑓 𝑐+ℎ −𝑓(𝑐)
ℎ
ℎ→𝑜
Hence lim
lim
≥0
This shows that right hand derivative 𝑓′(𝑐 + ) ≤ 0 and left hand derivative 𝑓′(𝑐 − ) ≥ 0.
Since 𝑓′(𝑥) exists at every value of 𝑥 in (a, b) and so 𝑓′ 𝑐 also exists. For this, above
two limits must exist, and equal to zero.
Hence 𝑓′ 𝑐 = 0 for some 𝑐 in (a, b).
Case III: If 𝑚 is different from 𝑓 𝑎 or 𝑓 𝑏 with 𝑀 ≠ 𝑚 we can similarly show that
𝑓′ 𝑐 = 0 for some 𝑐 in (a, b).
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Geometric interpretation
Let 𝑦 = 𝑓(𝑥) be a continuous function on the interval [a, b] and is differentiable
at each point in (a, b) such that 𝑓(𝑎) = 𝑓(𝑏). Then, there exists at least a point
𝑐 ∈ (𝑎, 𝑏) where the tangent drawn to the curve is parallel to x-axis.
i.e., 𝑓 ′(𝑐) = 0 as shown in figure.
This means that tangent is parallel to x-axis at the point x = c in the interval (a, b).
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Remark: 1
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1. Rolle’s theorem guarantees the existence of at least one point c such that
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𝑓′(𝑐) = 0. It does not restrict the existence of more than one point such
that the tangent is parallel to the x-axis.
2. If any of the conditions of Rolle's theorem fails for a given function, we
cannot apply Rolle's theorem.
For example, consider a function 𝑓(𝑥) = |𝑥|, 𝑥 ∈ [– 1, 1]. The function 𝑓(𝑥)
is continuous on [– 1, 1] and 𝑓(– 1) = 𝑓(1) = 1. But 𝑓′(0) does not exist at
𝑐 = 0 and c ∈ (– 1, 1). So, 𝑓(𝑥) is not differentiable at all points in (– 1, 1).
Hence, we can not apply Rolle's theorem in this example.
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Solution.
First of all, we need to check that the function f(x) satisfies all the conditions of
Rolle's theorem.
1.
f(x) is continuous in [−2,0] as a quadratic function;
2. It is differentiable everywhere over the open interval (−2,0);
3. Finally,
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Example 2.
Given the function 𝑓(𝑥) = 𝑥2 − 6𝑥 + 5. Find all values of c in the open
interval (2,4) such that 𝑓′(𝑐) = 0.
Solution.
First we determine whether Rolle's theorem can be applied to 𝑓(𝑥) on
the closed interval [2,4].
The function is continuous on the closed interval [2,4].
The function is differentiable on the open interval (2,4).
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Its derivatives
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Example 3.
Let 𝑓(𝑥) = 𝑥2 + 8𝑥 + 14. Find all values of c in the interval (−6,−2) such that 𝑓′(𝑐) = 0.
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Lagrange's Mean Value Theorem (MVT)
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Lagrange's mean value theorem (MVT) states that if a function 𝑓(𝑥) is continuous
on a closed interval [a, b] and differentiable on the open interval (a, b), then
there is at least one point 𝑥 = 𝑐 on this interval, such that
𝑓′
𝑓 𝑏 −𝑓 𝑎
𝑐 =
(𝑏 − 𝑎)
Proof: Let us define a function
𝐹 𝑥 = 𝑓 𝑥 + 𝐾𝑥
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………… (i)
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Where 𝑓(𝑥) and 𝐾𝑥 are continuous in 𝑎, 𝑏 and differentiable in (𝑎, 𝑏).
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therefore, it follows that the sum of continuous functions is continuous and sum
of differentiable functions is differentiable.
So, 𝐹 𝑥 is continuous in 𝑎, 𝑏 and differentiable in (𝑎, 𝑏).
The 𝐾 is constant to determined so that 𝐹 𝑎 = 𝐹 𝑏 .
i.e. ,
f 𝑎 + 𝐾𝑎 = 𝑓 𝑏 + 𝐾𝑏
or,
𝐾(𝑎– 𝑏) = 𝑓(𝑏) – 𝑓(𝑎)
∴
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−𝐾 =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
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Thus F(x) satisfies all three conditions of Rolle’s Theorem. Hence there exist at
least one point 𝑐 in 𝑎, 𝑏 such that
𝐹′ 𝑐 = 0
so from (i)
…….. (ii)
𝐹 𝑥 = 𝑓 𝑥 + 𝐾𝑥
Differentiate with respect to 𝑥
𝐹′ 𝑥 = 𝑓′ 𝑥 + 𝐾
or,
𝐹′ 𝑐 = 𝑓′ 𝑐 + 𝐾
From (ii)
𝑓′ 𝑐 = −𝐾
∴
𝑓′ 𝑐 =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
𝑎<𝑐<𝑏
Which proves the theorem
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Geometric interpretation of mean value theorem
Let 𝑓(𝑥) be continuous on [a, b] and differentiable
on (a, b) as shown in the figure.
Then the slope of the secant line joining A and B is
m = 𝑡𝑎𝑛𝜃 =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
Also, the slope of tangent at 𝑐 is 𝑓′(𝑐). We see from the graph that the tangent
drawn at C is parallel to secant line joining endpoints A and B, i.e.,
𝑓 ′(𝑐) =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
Remark: There may be more than one point in (a, b) satisfying Lagrange's mean
value theorem.
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Example 1. Check the validity of Lagrange's mean value theorem for the function 𝑓(𝑥) =
𝑥2 − 3𝑥 + 5 on the interval [1,4]. If the theorem holds, find a point c satisfying the conditions
of the theorem.
Solution. The given quadratic function is continuous and differentiable on the entire set of
real numbers. Hence, we can apply Lagrange's mean value theorem. The derivative of the
function has the form
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Any Questions…?
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Thank You VERY much
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