MEAN VALUE THEOREM
 STATEMENT Let f be a real valued function defined on
the closed interval [a, b] such that
 It is continuous on the closed interval [a, b].
 It is differentiable on the open interval (a, b).
Then, there exist a real number c∈ 𝑎, 𝑏
𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
𝑓′
𝑓 𝑏 −𝑓 𝑎
𝑐 =
.
𝑏−𝑎
The expression (f(b) − f(a)) / (b − a) gives the slope of the line joining
the points (a, f(a)) and (b, f(b)), which is a chord of the graph of f, while
f′(x) gives the slope of the tangent to the curve at the point (x, f(x)).
Thus the Mean value theorem says that given any chord of a smooth
curve, we can find a point lying between the end-points of the chord
such that the tangent at that point is parallel to the chord.
 Proof: Consider a function 𝜑 𝑥 = 𝑓 𝑥 +
𝐴𝑥, 𝑤ℎ𝑒𝑟𝑒 𝐴 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑏𝑒 𝑐ℎ𝑜𝑠𝑒𝑛 𝑖𝑛 𝑠𝑢𝑐ℎ 𝑎 𝑤𝑎𝑦 𝑡ℎ𝑎𝑡 𝜑 𝑎 = 𝜑 𝑏 .
 Now, Let 𝜑 𝑎 = 𝜑 𝑏
=> 𝑓 𝑎 + 𝐴𝑎 = 𝑓 𝑏 + 𝐴𝑏
 => 𝑓 𝑏 − 𝑓 𝑎 = −𝐴 𝑏 − 𝑎
 𝐴={







𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
}
Since f(x) is continuous on [a, b] and Ax, being a polynomial function, is everywhere
continuous. Therefore, 𝜑 𝑥 , being the sum of two continuous function f(x) and Ax, is
continuous on [a, b].
Since f(x) is differentiable on (a, b) and Ax, being a polynomial function, is everywhere
differentiable. Therefore, 𝜑 𝑥 , being the sum of two differentiable function f(x) and Ax,
is differentiable on (a, b).
𝜑 𝑎 =𝜑 𝑏 .
Thus, all the three conditions of Rolle’s theorem are satisfied by , 𝜑 𝑥 on [a, b]. so, there
must exist some 𝑐 ∈ 𝑎, 𝑏 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝜑′ (𝑐) = 0.
Now, 𝜑 𝑥 = 𝑓 ′ 𝑥 + 𝐴𝑥
𝜑 𝑥 = 𝑓′ 𝑥 + 𝐴
𝜑′ 𝑐 = 𝑓 ′ 𝑐 + 𝐴
 𝜑′ 𝑐 = 0
 f’(c)+A=0 =>f’(c)=-A
 f’(c) =
𝑓 𝑏 −𝑓 𝑎
𝑏−𝑎
Geometrical Interpretation
 let f(x) be a function defined on [a, b] and let APB be the curve
represented by y=f(x). then coordinate of A and B are (a, f(a) and
b, f(b)) respectively. Suppose the chord AB makes an angle 𝜓
with the axis of x. Then, from the triangle ARB, we have
 tan
𝐵𝑅
𝜓=
𝐴𝑅
=> 𝑡𝑎𝑛𝜓
𝑓 𝑏 −𝑓 𝑎
=
𝑏−𝑎
 by Lagrange’s Mean Value Theorem, we have
 f’(c)
𝑓 𝑏 −𝑓 𝑎
=
𝑏−𝑎

 => 𝑡𝑎𝑛𝜓 = 𝑓 ′ 𝑐
 => 𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑜𝑟𝑑 𝐴𝐵 = 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑎𝑡 (𝑐, 𝑓(𝑐))
Working Rule
 We first check whether f(x) satisfies conditions of Mean Value Theorem or not.
 Differentiate the function to get f’(c).
𝒇 𝒃 −𝒇 𝒂
 Equate the derivatives with 𝒇′ 𝒄 =
. to get the value of c.
𝒃−𝒂
 If c belongs to given interval then, Mean Value Theorem verified.
Note:
 The following rules may be helpful while solving the problem.
 A polynomial function is everywhere continuous and differentiable
 The exponential function, sine and cosine function are everywhere continuous and
differentiable.
 Logarithmic function is continuous and differentiable in its domain.
𝝅
𝟐
 Tan x is not continuous at x = ± , ±

𝟑𝝅
𝟓𝝅
,± …….
𝟐
𝟐
𝒙 𝒊𝒔 𝒏𝒐𝒕 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒃𝒍𝒆 𝒂𝒕 𝒙 = 𝟎.
 𝑖𝑓 𝑓 ′ 𝑥 𝑡𝑒𝑛𝑑𝑠 𝑡𝑜 ± ∞ 𝑎𝑠 𝑥 → 𝑘, 𝑡ℎ𝑒𝑛 𝑓 𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 𝑘.
Verify Lagrange’s mean value theorem for the function
f(x)=(x-3)(x-6)(x-9) on the interval [3, 5].





f(x)=(x-3)(x-6)(x-9) =x3 -18x2 + 99x -162.
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [3, 5] and differentiable on (3, 5).
Thus, both the conditions of Lagrange’s mean value theorem are satisfied.
So, there must exist at least one real number c ∈ 𝑎, 𝑏 such that
𝑓 5 −𝑓(3)
 f’(c)=
5−3
 Now, f(x) = x3 -18x2 + 99x -162
 f’(x)=3x2 -36x+99, f(5)=(5-3)(5-6)(5-9)=8 and f(3)=0
𝑓 5 −𝑓(3)
5−3
8−0
3x2 -36x+99=5−3
 f’(x)=

 3x2 -36x+99=4
 3x2 -36x+95=0
 =>
36± 1296−1140
6
=
36±12.49
6
= 8.8, 4.8
𝑓 5 −𝑓(3)
 Thus, c=4.8∈(3, 5) such that f’(c) =
.
5−3
 Hence, Lagrange’s mean value theorem is verified.
Verify Lagrange’s mean value theorem for the following function on the
indicated intervals. Also, find a point c in the indicated interval.
F(x)=x(x-2) on [1, 3]




f(x)=x(x-2)
Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [1, 3] and differentiable on (1, 3).
Thus, both the conditions of Lagrange’s mean value theorem are satisfied on [1,
3].
 So, there must exist at least one real number c ∈ 1, 3 such that
𝑓 3 −𝑓(1)
 f’(c)=
3−1
 Now, f(x) = x2 -2x
 f’(x) =2x-2, f(3)=9-6=3 and f(1)=1-2=-1
𝑓 3 −𝑓(1)
3−1
3 − −1
2x-2= 3−1
 f’(x)=

 2x-2=2
=>x=2
𝑓 3 −𝑓(1)
 Thus, c=2∈(1, 3) such that f’(c)=
.
3−1
 Hence, Lagrange’s mean value theorem is verified.
Verify Lagrange’s mean value theorem for the following
function on the indicated intervals. Also, find a point c
in the indicated interval. f(x) = x (x-1) (x-2) on interval [0, ½]




Since a polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [0, 1/2] and differentiable on ( 1/2).
Thus, both the conditions of Lagrange’s mean value theorem are satisfied on [0, 1/2].
So, there must exist at least one real number c ∈ 0, 1/2 such that
 f’(c)=
𝑓 1/2 −𝑓(0)
1/2−0
 Now, f(x)=x3-3x2+2x.
 => f’(x)=3x2-6x+2, f(0)=0 and f(1/2)=1/8-3/4+1=3/8.
𝑓 1/2 −𝑓(0)
1/2−0
3/8 −(0)
3x2-6x+2 =
1/2−0
3
3x2-6x+2=
4
 F’(x )=


 12x2-24x+5=0
 => x=
24± 336
24
21
c=16
=1±
21
6
𝑓 1/2 −𝑓(0)
∈ [0, ½] such that f’(c)=
1/2−0
 Hence, Lagrange’s mean value theorem is verified.
 Since,
1
𝑥
 Example. Let , 𝑓 𝑥 = , a = -1and b=1.
 We have


𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
2
2
= =1
 On the other hand, for any 𝑐 ∈ (−1, 1), not equal to 0, we have
 𝑓′ 𝑐 = −
1
𝑐2
≠ 1.

 So the equation 𝑓 ′ 𝑐 =
𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
 does not have a solution in c. This does not contradict the Mean Value Theorem, since
f(x) is not even continuous on [-1,1].
 Remark. It is clear that the derivative of a constant function is 0. But you may wonder
whether a function with derivative zero is constant. The answer is yes. Indeed, let f(x) be
a differentiable function on an interval I, with f '(x) =0, for every 𝑥 ∈ I. Then for any a and
b in I, the Mean Value Theorem implies

𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
= 𝑓′(𝑐)

 for some c between a and b. So our assumption implies 𝑓 𝑏 − 𝑓 𝑎 = 0. 𝑏 − 𝑎 = 0.


 Thus f(b) = f(a) for any aand b in I, which means that f(x) is constant.
 Exercise 1. Show that the equation 2x3 + 3x2 + 6x + 1 = 0 has exactly one real
root.
 Answer.
 Let f(x) = 2x3 + 3x2 + 6x + 1. We have f(0)=1 and f(-1) = -4. So the Intermediate
Value Theorem shows that there exists a point c between -1 and 0 such that f(c)
=0. Consequently our equation has at least one real root.
 Let us now show that this equation has also at most one real root. Assume not,
then there must exist at least two roots c1and c2, with c1 < c2. Then we have f(c1)
=0 and f(c2) =0. Rolle's Theorem implies the existence of a point c between c1
and c2 such that
 f'’(c) = 6c2 + 6c + 6 =0.

 But the quadratic equation 6c2 + 6c + 6 =0 does not have real roots, yielding a
contradiction to our assumption that f(x) had at least two roots. Conclusion:
our original equation has exactly one real root.
 Exercise 2. Show that cos 2𝑎 − cos(2𝑏) ≤ 2 𝑎 − 𝑏
for all real numbers a and b.
 Answer. Let 𝑓 𝑥 = cos(2𝑥). Then we have𝑓 ′ 𝑥 =
− 2sin(2𝑥) . The inequality clearly holds when a=b.
For any numbers a and b, with𝑎 ≠ 𝑏 , the Mean Value
Theorem implies

𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
=
cos 2𝑏 −cos(2𝑎)
𝑏−𝑎
= −2sin(2𝑐)
 𝑓 𝑏 − 𝑓 𝑎 = −2 𝑏 − 𝑎 𝑠𝑖𝑛2𝑐
 𝑐𝑜𝑠2𝑏 − 𝑐𝑜𝑠2𝑎 = −2 𝑏 − 𝑎 𝑠𝑖𝑛2𝑐
 𝑐𝑜𝑠2𝑏 − 𝑐𝑜𝑠2𝑎 = 2 𝑏 − 𝑎
 𝑐𝑜𝑠2𝑏 − 𝑐𝑜𝑠2𝑎 ≤ 2 𝑏 − 𝑎
𝑠𝑖𝑛2𝑐 ≤ 1(proved)
𝑠𝑖𝑛2𝑐
𝑠𝑖𝑛𝑐𝑒 0 ≤
Exercise
 f(x)=x(x-1) on [1, 2]
 f(x)= x2 -3x +2 on [-1, 2]
 f(x)= x+1/x on [1, 3]
 f(x)= 𝑥 2 − 4 𝑜𝑛 [2, 4]
 f(x)=x2 +x +1 on [0, 4].
 Find the point on the curve y=x3 -3x, where the tangent
is parallel to the chord joining (1, -2) and (2, 2).