BCA 21 / IMCA 21 / Mathematics SVT 87
MEAN VALUE THEOREMS
Rolle's Theorem
Statement : If f (x) is a function
(i) Continuous in the closed interval a ≤ x ≤ b
(ii) differentiable in the open interval a < x < b and
(iii ) f (a ) = f (b) then there exists at least one value c of x such that f ′(c) = 0 for a < c < b.
P
Geometrical Meaning y = f (x )
A & B are points on the curve such that f (a ) = f (b). Tangent at P is parallel to A B
AB such that the slope of the tangent is f ′(c) = 0
f (a ) f (b )
Lagrange's Mean Value Theorem a c b
Statement : If f (x) is a function
(i) Continuous in the closed interval a ≤ x ≤ b
(ii) differentiable in the open interval a < x < b
f (b) − f (a)
then there exists at least one value c of x in the interval such that = f ′(c)
b−a
Note : - If the interval is (a, a + h) then f (a + h) − f (a ) = f ′(c)
ie f (a + h) = f (a ) + hf ′(c)
P
Geometrical Meaning
B
A & B are points on the curve corresponding to x = a & x = b. Join AB There
A
will be a tangent at P, parallel to AB so that scope of the tangent is f (b )
f (a )
f (b ) − f ( a )
which is f ′(c) a c b
b−a
Cauchy's Mean Value Theorem
Statement : If f ( x) & g ( x) are two functions which are
(i) continuous in the closed interval [a, b]
(ii) differentiable in the open interval (a, b) and
(iii) g ′( x) ≠ 0 for any value of x in (a, b)
f (b) − f (a ) f ′(c)
then there exists at least one value c of x in (a, b) such that =
g (b) − g (a ) g ′(c)
Taylor's Theorem
If f ( x) is a function such that (i) f ( x) and its (n − 1) derivatives are continuous in the closed interval [a, a + h]
ie a ≤ x ≤ a + h and (ii) n th derivative f ( n) ( x) exists in the open interval (a, a + h).
, 88 KSOU Differential Calculus
Then there exists at least one number θ (0 < θ < 1) such that
h2 h n ( n)
f (a + h) = f (a) + hf ′(a) + f ′′(a) + L + L + f (a + θh)
21 n!
x2 x n ( n)
Note : - put a = 0 & h = x, then f ( x) = f (0) + xf ′(0) + f ′′(0) + L + L + f (θx), where 0 < θ < 1
21 n!
xn
This expression for f ( x) is called Maclaurin' s Expansion and further if f (θx) tends to zero as n → ∞.
n!
x2
Then f ( x) = f (0) + xf ′(0) + f ′′(0) + LL to ∞. This is called Maclaurin' s Series for f ( x).
2!
Examples
(1) Verify Rolle' s Theorem for f ( x) = ( x + 2) 3 ( x − 3) 4 in (−2, 3) and find c.
Solution : f ( x) is continuous in the closed interval [−2, 3] & differentiable in (−2, 3) and further f (−2) = 0, f (3) = 0
ie f (−2) = f (3)
∴ There exists a value c in (−2, 3) such that f ′(c) = 0
f ′( x) = 3( x + 2) 2 ( x − 3) 4 + 4( x + 2) 3 ( x − 3) 3
f ′(c) = 3(c + 2) 2 (c − 3) 4 + 4(c + 2) 3 (c − 3) 3 = (c + 2) 2 (c − 3) 3 [3(c − 3) + 4(c + 2)] = (c + 2) 2 (c − 3) 2 (7c − 1)
1
f ′(c) = 0 ⇒ 7c − 1 = 0 ⇒ c =
7
( 2) Find ' c' of the Lagrange' s Mean Value Theorem for the function f ( x ) = ( x − 1)( x − 2)( x − 3) in (0, 4)
Solution : f (0) = (−1)(−2)(−3) = −6, f (4) = 3 × 2 × 1 = 6
f ′( x) = ( x − 2)( x − 3) + ( x − 1)( x − 3) + ( x − 1)( x − 2)
f ′(c) = (c − 2)(c − 3) + (c − 1)(c − 3) + (c − 1)(c − 2) = c 2 − 5c + b + c 2 − 4c + 3 + c 2 − 3c + 2 = 3c 2 − 12c + 11
f (b) − f (a )
By Lagrange' s Theorem = f ′(c)
b−a
f ( 4) − f ( 0)
∴ = f ′(c)
4−0
6+6
∴ = 3c 2 − 12c + 11
4
ie 12c 2 − 48c + 44 = 12
ie 12c 2 − 48c + 32 = 0 ⇒ 3c 2 − 12c + 8 = 0
12 ± 144 − 96 12 ± 48 12 ± 4 3
∴ c= = =
6 6 6
2 2 2
∴ c = 2± ie c = 2 − & 2+
3 3 3
MEAN VALUE THEOREMS
Rolle's Theorem
Statement : If f (x) is a function
(i) Continuous in the closed interval a ≤ x ≤ b
(ii) differentiable in the open interval a < x < b and
(iii ) f (a ) = f (b) then there exists at least one value c of x such that f ′(c) = 0 for a < c < b.
P
Geometrical Meaning y = f (x )
A & B are points on the curve such that f (a ) = f (b). Tangent at P is parallel to A B
AB such that the slope of the tangent is f ′(c) = 0
f (a ) f (b )
Lagrange's Mean Value Theorem a c b
Statement : If f (x) is a function
(i) Continuous in the closed interval a ≤ x ≤ b
(ii) differentiable in the open interval a < x < b
f (b) − f (a)
then there exists at least one value c of x in the interval such that = f ′(c)
b−a
Note : - If the interval is (a, a + h) then f (a + h) − f (a ) = f ′(c)
ie f (a + h) = f (a ) + hf ′(c)
P
Geometrical Meaning
B
A & B are points on the curve corresponding to x = a & x = b. Join AB There
A
will be a tangent at P, parallel to AB so that scope of the tangent is f (b )
f (a )
f (b ) − f ( a )
which is f ′(c) a c b
b−a
Cauchy's Mean Value Theorem
Statement : If f ( x) & g ( x) are two functions which are
(i) continuous in the closed interval [a, b]
(ii) differentiable in the open interval (a, b) and
(iii) g ′( x) ≠ 0 for any value of x in (a, b)
f (b) − f (a ) f ′(c)
then there exists at least one value c of x in (a, b) such that =
g (b) − g (a ) g ′(c)
Taylor's Theorem
If f ( x) is a function such that (i) f ( x) and its (n − 1) derivatives are continuous in the closed interval [a, a + h]
ie a ≤ x ≤ a + h and (ii) n th derivative f ( n) ( x) exists in the open interval (a, a + h).
, 88 KSOU Differential Calculus
Then there exists at least one number θ (0 < θ < 1) such that
h2 h n ( n)
f (a + h) = f (a) + hf ′(a) + f ′′(a) + L + L + f (a + θh)
21 n!
x2 x n ( n)
Note : - put a = 0 & h = x, then f ( x) = f (0) + xf ′(0) + f ′′(0) + L + L + f (θx), where 0 < θ < 1
21 n!
xn
This expression for f ( x) is called Maclaurin' s Expansion and further if f (θx) tends to zero as n → ∞.
n!
x2
Then f ( x) = f (0) + xf ′(0) + f ′′(0) + LL to ∞. This is called Maclaurin' s Series for f ( x).
2!
Examples
(1) Verify Rolle' s Theorem for f ( x) = ( x + 2) 3 ( x − 3) 4 in (−2, 3) and find c.
Solution : f ( x) is continuous in the closed interval [−2, 3] & differentiable in (−2, 3) and further f (−2) = 0, f (3) = 0
ie f (−2) = f (3)
∴ There exists a value c in (−2, 3) such that f ′(c) = 0
f ′( x) = 3( x + 2) 2 ( x − 3) 4 + 4( x + 2) 3 ( x − 3) 3
f ′(c) = 3(c + 2) 2 (c − 3) 4 + 4(c + 2) 3 (c − 3) 3 = (c + 2) 2 (c − 3) 3 [3(c − 3) + 4(c + 2)] = (c + 2) 2 (c − 3) 2 (7c − 1)
1
f ′(c) = 0 ⇒ 7c − 1 = 0 ⇒ c =
7
( 2) Find ' c' of the Lagrange' s Mean Value Theorem for the function f ( x ) = ( x − 1)( x − 2)( x − 3) in (0, 4)
Solution : f (0) = (−1)(−2)(−3) = −6, f (4) = 3 × 2 × 1 = 6
f ′( x) = ( x − 2)( x − 3) + ( x − 1)( x − 3) + ( x − 1)( x − 2)
f ′(c) = (c − 2)(c − 3) + (c − 1)(c − 3) + (c − 1)(c − 2) = c 2 − 5c + b + c 2 − 4c + 3 + c 2 − 3c + 2 = 3c 2 − 12c + 11
f (b) − f (a )
By Lagrange' s Theorem = f ′(c)
b−a
f ( 4) − f ( 0)
∴ = f ′(c)
4−0
6+6
∴ = 3c 2 − 12c + 11
4
ie 12c 2 − 48c + 44 = 12
ie 12c 2 − 48c + 32 = 0 ⇒ 3c 2 − 12c + 8 = 0
12 ± 144 − 96 12 ± 48 12 ± 4 3
∴ c= = =
6 6 6
2 2 2
∴ c = 2± ie c = 2 − & 2+
3 3 3
