1. Sameem Makhdoomi-09
Suhaib Hamdani-17
Tariq Hameed-03

2. CAUCHY MEAN VALUE THEOREM
Statement:
If f(x) and g(x) are two functions and any interval a
to b is given then conditions for satisfying Cauchy
Mean Value Theorem are:
1) f(x) and g(x) be continuous in closed interval
[a b].
2) Differentiable in open interval (a b).
i.e. f’(x) and g’(x) exists in (a b).
3) And g’(x) != 0 for any value of x belong to (a b)
Presentation by Sameem, Tariq and Suhaib

3. Closed interval and Open interval
1. Close interval [a b]:
Checking continuity of any function f(x) and
g(x) in an interval which lie between points a
and b *including end points a and b.*
2. Open interval (a b):
Checking continuity of any function f(x) and
g(x) in an interval which lie between points a
and b *excluding end points a and b.*
Presentation by Sameem, Tariq and Suhaib

4. Then there exist a value ‘C’ in between the
interval (a b) , such that:
f’(c) / g’(c) = f(b)-f(a) / g(b)-g(a);
f(a) C f(b)
Presentation by Sameem, Tariq and Suhaib

5. Proof of Cauchy Mean Value
Theorem
Let F(x) be a function, such that:
F(x)=f(x)+Ag(x) --- (1)
where A is constant to be determine such that,
F(a)=F(b)
Replace ‘x’ by ‘a’ in equation (1),
F(a)=f(a)+g(b)A ---(2)
F(b)=f(b)+g(b)A ----(3)
F(a)=F(b) => f(a)+g(a)A=f(b)+g(b)A
Ag(a)-Ag(b)=f(b)-f(a)
Presentation by Sameem, Tariq and Suhaib

6. -A{g(b)-g(a)}=f(b)-f(a)
-A=f(b)-f(a)/g(b)-g(a) -----(4)
Since g(x) and f(x) both are continuous and
differentiable so F(x) is also continuous and
differentiable
By Rolls Theorem
F’(c)=0
F(x)=f(x)+g(x)A
F’(x)=f’(x)+g’(x)A
x=c;
F’(c)=f’(c)+g’(c)A
Presentation by Sameem, Tariq and Suhaib

7. Presentation by Sameem, Tariq and Suhaib

8. Q1) Verify CMVT for the function x^2 and
x^3 in [1,2].
Sol.
Let, f(x) =x^2 and f’(x)= 2x. a=1
g(x)= x^3 and g’(x)= 3x^2 b=2
1. Clearly f(x) and g(x) are continuous in [1,2]
and differentiable in (1,2).
2. And also g’(x)=3x^2 will not be equal to 0
in an interval (1,2).
Then by CMVT, there exist a point ‘c’ which
belongs to (1,2) such that,
f’(c)/g’(c)= f(b) -f (a) / g(b) - g(a)
Presentation by Sameem, Tariq and Suhaib

9. f’(c)/g’(c)=f(2)-f(1)/g(2)-g(1)
[ f(x)=x^2 and g(x)=x^3]
2c/3c^2=4-1/8-1
2/3c=3/7
c =14/9 =1.55
Therefore the value of point c (here i.e 1.55)
lies between (1,2).
Hence C.M.V.T is verified.
Presentation by Sameem, Tariq and Suhaib

10. Q2): Find the value of “C” using CMVT for
the functions e^x and e^-x in an interval
(0,3)?
Sol:
Let, f(x)=e^x and f’(x)=e^x. a=0
g(x)=e^-x and g’(x)=-e^-x. b=3
[ Both are Continuous In an interval [0,3] and
Differentiable In an interval (0,3) and also
g’(x) = -e^-x ]
Presentation by Sameem, Tariq and Suhaib

11. By CMVT
Then there exist point ‘c’ in an interval (0,3) such
that,
f’(c)/g’(c)= f(b) -f (a) / g(b) - g(a)
f’(c)/g’(c)=f(3)-f(0)/g(3)-g(0)
e^c / e^-c =e^3 – e^0 / e^-3 – e^0
e^c / -(1/e^c) = (e^3 –1) / (1/e^3 -1)
(e^3 -1) / (1- e^3 / e^3) = -e^2c
[(e^3 -1)* e^3] /- (e^3 -1)= -e^2c
e^3=e^2c
3=2c
c=3/2=1.5
Presentation by Sameem, Tariq and Suhaib

12. That’s all from our side
Thankyou.