Unit 2

Mean value theorems

Continuity- suppose that a function f(x) is defined in the interval I , then it is said to be continuous at x=a , if

Differentiability- A function f(x) is said to be differentiable at x=a if

exists where ‘a’ belongs to I

Rolle’s theorem-

Suppose f(x) is a function defined on [a , b] and it satisfies the following conditions

1. f(x) is continuous in [a , b]

2. f(x) is differentiable in (a , b)

3. f(a) = f(b)

Then there exists atleast a point point c ϵ (a , b) , where a<b , such that f’(c) = 0

Proof: suppose y = f(x) is a function and A(a , f(a)) , B(b , f(b)) be two points on the curve f(x) and a,b are two end points. Now conditions for Rolle’s theorem-

1.f(x) is a continuous function in[a , b] , from the figure without breaks in between A&B on y = f(x).

2. f(x) is differentiable in (a , b), because joining A and B we get a line AB.

Slope of the line AB=0 then a point C at P also a tangent at P, or Q,R,S is parallel to x –axis.

Slope of the tangent at P or Q,R, S, will be 0,even the curve y = f(x) decreases or increases, that means f(x) is constant.

Derivative of f(x),

f’(c) = 0

That’s why, f’(c) = 0

3. The slope of the line AB is equal to zero, that means the line AB is parallel to x-axis.

So that, f(a) = f(b)

Example 1: Verify Rolle’s theorem for the function f(x) = x(x+3) in interval [-3, 0].

Solution: First we will differentiate the given function with respect to x, we get

f’(x) = (x²+3x) + (2x + 3)

=

This shows that f’(x) exists for all x, therefore f(x) is continuous for all x.

Now, f(-3) = 0 and f(0) = 0 , so that f(-3) = f(0).

Here f(x) satisfies all the conditions of Rolle’s theorem,

Then,

f’(x) = 0 , which gives

= 0

We get,

X = 3 and x = -2

Here we can see that clearly -3<-2<0 , therefore there exists -2 ∈ (-3,0) such that

f’(-2) = 0

That means the Rolle’s theorem is true for the given function.

Example 2: Verify Rolle’s theorem for the given functions below-

1. f(x) = x³ - 6x²+11x-6 in the interval [1,3]

2. f(x) = x²-4x+8 in the interval [1,3]

Solution. (1) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = f(3) = 0

Now we find f’(x) = 0

3x² - 12x +11 = 0

We get, x = 2+ and 2 -

Hence both of them lie in (1,3).

Hence the theorem holds good for the given function in interval [1,3]

(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = 1 -4 +8 = 5 and f(3) = 9 – 12 + 8 = 5

Hence f(1) = f(3)

Now the first derivative of the function,

f’(x) = 0

2x – 4 = 0 , gives

X = 2

We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0.

This means that the Rolle’s theorem holds good for the given function and given interval.

Example 3: Verify the Rolle’s theorem for sin x in the interval []

Solution: Suppose f(x) = sin x

We know that sin x is continuous for all x.

Now, f’(x) = cos x exists for all x in () and

f(0

f(0

Thus f(x) satisfies all the conditions of Rolle’s theorem.

Now,

f’(x) = 0 that gives, cos x = 0

x =

Here we notice that both intervals lie in (.

There exists, c =

So that, f’(c) = 0

The Rolle’s theorem has been verified.

Suppose that f(x) be a function of x such that,

1. If it is continuous in [a , b]

2. If it is differentiable in (a , b)

Then there atleast exists a value cϵ (a , b)

f’(c) =

Proof:

Lets define a function g(x),

g(x) = f(x) – Ax ………………..(1)

Here A is a constant which is to be determined,

So that, g(a) = g(b)

Now,

g(a) = f(a) – Aa

g(b) = f(b) – Ab

So,

g(a) = g(b),

f(a) – Aa = f(b) – Ab ,

Which gives,

A = …………………..(2)

As right hand side of eq.(1) is continuous in [a,b] , so that g(x) is continuous.

And right hand side of eq.(1) is differential in (a,b) , so that g(x) is differentiable in (a,b).

And g(a) = g(b) , because of the choice of A.

Hence g(x) satisfies all the conditions of Rolle’s theorem.

So that,

There exists a value c such that a<c<b at which g’(c) = 0

Now, differentiate eq. (1) with respect to x, we get

g’(x) = f’(x) – A

Here we know that, x = c,

g’(c) = f’(c) – A

As g’(c) = 0, then

f’(c) – A =0

So that,, f’(c) = A,

From equation (2) , we get

f’(c) = hence proved.

Example 1: Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].

Solution: As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4).

f(x) = (x-1)(x-2)(x-3)

f(x) = x-6x²+11x-6

Now at x = 0, we get

f(0) = -6 and

At x = 4, we get.

f(4) = 6

Diff. The function w.r.t.x , we get

f’(x) = 3x²-6x+11

Suppose x = c, we get

f’(c) = 3c²-6c+11

By Lagrange’s mean value theorem,

f’(c) = = = = 3

Now we get,

3c²-6c+11 = 3

3c²-6c+8 = 0

On solving the quadratic equation, we get

C = 2

Here we see that the value of c lies between 0 and 4

Therefore, the given function is verified.

Example 2: Verify Lagrange’s mean value theorem for f(x) = log xin [1,e].

Solution: We already know that the function which is log x is continuous for all x>0.

So that this is the continuous function In [1,e]

Now,

f’(x) = 1/x

Which is exists for all x in (1,e)

So that f(x) is differentiable in (1,e).

By Lagrange’s mean value theorem, we get

f’(c) = , let x = c,

Then ,

f’(c) =

We get,

c = e-1

e-1 will always lies between 1 and e .

Hence the function is verified by Lagrange’s mean value theorem.

Suppose we have two functions f(x) and g(x) of x, such that,

1. Both functions are continuous in [a,b]

2. Both functions are differentiable in (a,b)

3. g’(x) ≠ 0 for any x ϵ (a,b)

These three exists atleast , x = c ϵ (a,b) , at which

Proof: suppose, we define a functions,

h(x) = f(x) – A.g(x) …………………….(1)

So that h(a) = h(b) and A is a constant to be determined.

Now,

h(a) = f(a) – Ag(a)

h(b) = f(b) – A.g(b)

So that,

f(a) – Ag(a) = f(b) – A.g(b) , which gives

A = …………………………….(2)

Now, h(x) is continuous in [a,b] as RHS of eq. (1) is continuous in [a,b] and h(x) is diff. In (a,b) as RHS of eq. (1) is diff. In (a,b)

Also,

h(a) = h(b)

Therefore, all the conditions of Rolle’s theorem are satisfied then there exists a Value x = cϵ (a, b)

So that h’(c) = 0

Differentiate eq.(1) w.r.t.x, we get

h’(x) = f’(x) – A.g’(x)

At x = c

h’(c) = f’(c) – A.g’(c)

0 = f’(c) – A.g’(c)

A =

So that , we get

where a<c<b

Hence the Cauchy’s mean value theorem is proved.

Example 1: Verify Cauchy’s mean value theorem for the function f(x) = sin x and g(x) = cosx in [ 0, π/2]

Solution: It is given the,

f(x) = sin x and g(x) = cos x

Now,

f’(x) = cos x and g’(x) = - sin x

We know that both the functions are continuous in [ 0 , π/2] and differentiable in ( 0 , π/2 )

Also, g’(x) = -sin x ≠ 0 for all x ϵ( 0 , π/2 )

By Cauchy’s mean value theorem, we get

for some c: 0< c <

That means

which gives,

Cot c = 1

C =

Now we see that lies between 0 and

Example 2: Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]

Solution: We are given, f(x) = x⁴ and g(x) = x

Derivative of these functions,

f’(x) = 4x³ and g’(x) = 2x

Put these values in Cauchy’s formula, we get

2c² =

c² =

c =

Now put the values of a = 1 and b = 2 ,we get

c = = = (approx)

Hence the Cauchy’s theorem is verified.

Taylor’s Theorem-

If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-

+ …….+ + ……..

Which is called Taylor’s theorem.

If we put x = a, we get-

+ …….+ + …….. (1)

Maclaurin’s Theorem-

If we put a = 0 and h = x then equation(1) becomes-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)

We get-

+ …….

Example 1: Express the polynomial in powers of (x-2).

Solution: Here we have,

f(x) =

Differentiating the function w.r.t.x-

f’(x) =

f’’(x) = 12x + 14

f’’’(x) = 12

f’’’’(x)=0

Now using Taylor’s theorem-

+ ……. (1)

Here we have, a = 2,

Put x = 2 in the derivatives of f(x), we get-

f(2) =

f’(2) =

f’’(2) = 12(2)+14 = 38

f’’’(2) = 12 and f’’’’(2) = 0

Now put a = 2 and substitute the above values in equation (1), we get-

Example 2: Expand sin x in powers of

Solution: Let f(x) = sin x

Then,

=

By using Taylor’s theorem-

+ ……. (1)

Here f(x) = sin x and a = π/2

f’(x) = cosx, f’’(x) = - sin x, f’’’(x) = - cos x and so on.

Putting x = π/2, we get

f(x) = sin x = = 1

f’(x) = cos x = = 0

f’’(x) = -sin x = = -1

f’’’(x) = -cos x = = 0

From equation (1) put a = and substitute these values, we get-

+ …….

= ………………………..

Textbooks:

1. Erwin Kreyszig, Advanced Engineering Mathematics, 10/e, John Wiley & Sons, 2011.

2. B. S. Grewal, Higher Engineering Mathematics, 44/e, Khanna Publishers, 2017.

References:

1. R. K. Jain and S. R. K. Iyengar, Advanced Engineering Mathematics, 3/e, Alpha Science

International Ltd., 2002.

2. George B. Thomas, Maurice D. Weir and Joel Hass, Thomas Calculus, 13/e, Pearson

Publishers, 2013.

3. Glyn James, Advanced Modern Engineering Mathematics, 4/e, Pearson publishers, 201.

Unit 2

Mean value theorems

Continuity- suppose that a function f(x) is defined in the interval I , then it is said to be continuous at x=a , if

Differentiability- A function f(x) is said to be differentiable at x=a if

exists where ‘a’ belongs to I

Rolle’s theorem-

Suppose f(x) is a function defined on [a , b] and it satisfies the following conditions

1. f(x) is continuous in [a , b]

2. f(x) is differentiable in (a , b)

3. f(a) = f(b)

Then there exists atleast a point point c ϵ (a , b) , where a<b , such that f’(c) = 0

Proof: suppose y = f(x) is a function and A(a , f(a)) , B(b , f(b)) be two points on the curve f(x) and a,b are two end points. Now conditions for Rolle’s theorem-

1.f(x) is a continuous function in[a , b] , from the figure without breaks in between A&B on y = f(x).

2. f(x) is differentiable in (a , b), because joining A and B we get a line AB.

Slope of the line AB=0 then a point C at P also a tangent at P, or Q,R,S is parallel to x –axis.

Slope of the tangent at P or Q,R, S, will be 0,even the curve y = f(x) decreases or increases, that means f(x) is constant.

Derivative of f(x),

f’(c) = 0

That’s why, f’(c) = 0

3. The slope of the line AB is equal to zero, that means the line AB is parallel to x-axis.

So that, f(a) = f(b)

Example 1: Verify Rolle’s theorem for the function f(x) = x(x+3) in interval [-3, 0].

Solution: First we will differentiate the given function with respect to x, we get

f’(x) = (x²+3x) + (2x + 3)

=

This shows that f’(x) exists for all x, therefore f(x) is continuous for all x.

Now, f(-3) = 0 and f(0) = 0 , so that f(-3) = f(0).

Here f(x) satisfies all the conditions of Rolle’s theorem,

Then,

f’(x) = 0 , which gives

= 0

We get,

X = 3 and x = -2

Here we can see that clearly -3<-2<0 , therefore there exists -2 ∈ (-3,0) such that

f’(-2) = 0

That means the Rolle’s theorem is true for the given function.

Example 2: Verify Rolle’s theorem for the given functions below-

1. f(x) = x³ - 6x²+11x-6 in the interval [1,3]

2. f(x) = x²-4x+8 in the interval [1,3]

Solution. (1) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = f(3) = 0

Now we find f’(x) = 0

3x² - 12x +11 = 0

We get, x = 2+ and 2 -

Hence both of them lie in (1,3).

Hence the theorem holds good for the given function in interval [1,3]

(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = 1 -4 +8 = 5 and f(3) = 9 – 12 + 8 = 5

Hence f(1) = f(3)

Now the first derivative of the function,

f’(x) = 0

2x – 4 = 0 , gives

X = 2

We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0.

This means that the Rolle’s theorem holds good for the given function and given interval.

Example 3: Verify the Rolle’s theorem for sin x in the interval []

Solution: Suppose f(x) = sin x

We know that sin x is continuous for all x.

Now, f’(x) = cos x exists for all x in () and

f(0

f(0

Thus f(x) satisfies all the conditions of Rolle’s theorem.

Now,

f’(x) = 0 that gives, cos x = 0

x =

Here we notice that both intervals lie in (.

There exists, c =

So that, f’(c) = 0

The Rolle’s theorem has been verified.

Suppose that f(x) be a function of x such that,

1. If it is continuous in [a , b]

2. If it is differentiable in (a , b)

Then there atleast exists a value cϵ (a , b)

f’(c) =

Proof:

Lets define a function g(x),

g(x) = f(x) – Ax ………………..(1)

Here A is a constant which is to be determined,

So that, g(a) = g(b)

Now,

g(a) = f(a) – Aa

g(b) = f(b) – Ab

So,

g(a) = g(b),

f(a) – Aa = f(b) – Ab ,

Which gives,

A = …………………..(2)

As right hand side of eq.(1) is continuous in [a,b] , so that g(x) is continuous.

And right hand side of eq.(1) is differential in (a,b) , so that g(x) is differentiable in (a,b).

And g(a) = g(b) , because of the choice of A.

Hence g(x) satisfies all the conditions of Rolle’s theorem.

So that,

There exists a value c such that a<c<b at which g’(c) = 0

Now, differentiate eq. (1) with respect to x, we get

g’(x) = f’(x) – A

Here we know that, x = c,

g’(c) = f’(c) – A

As g’(c) = 0, then

f’(c) – A =0

So that,, f’(c) = A,

From equation (2) , we get

f’(c) = hence proved.

Example 1: Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].

Solution: As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4).

f(x) = (x-1)(x-2)(x-3)

f(x) = x-6x²+11x-6

Now at x = 0, we get

f(0) = -6 and

At x = 4, we get.

f(4) = 6

Diff. The function w.r.t.x , we get

f’(x) = 3x²-6x+11

Suppose x = c, we get

f’(c) = 3c²-6c+11

By Lagrange’s mean value theorem,

f’(c) = = = = 3

Now we get,

3c²-6c+11 = 3

3c²-6c+8 = 0

On solving the quadratic equation, we get

C = 2

Here we see that the value of c lies between 0 and 4

Therefore, the given function is verified.

Example 2: Verify Lagrange’s mean value theorem for f(x) = log xin [1,e].

Solution: We already know that the function which is log x is continuous for all x>0.

So that this is the continuous function In [1,e]

Now,

f’(x) = 1/x

Which is exists for all x in (1,e)

So that f(x) is differentiable in (1,e).

By Lagrange’s mean value theorem, we get

f’(c) = , let x = c,

Then ,

f’(c) =

We get,

c = e-1

e-1 will always lies between 1 and e .

Hence the function is verified by Lagrange’s mean value theorem.

Suppose we have two functions f(x) and g(x) of x, such that,

1. Both functions are continuous in [a,b]

2. Both functions are differentiable in (a,b)

3. g’(x) ≠ 0 for any x ϵ (a,b)

These three exists atleast , x = c ϵ (a,b) , at which

Proof: suppose, we define a functions,

h(x) = f(x) – A.g(x) …………………….(1)

So that h(a) = h(b) and A is a constant to be determined.

Now,

h(a) = f(a) – Ag(a)

h(b) = f(b) – A.g(b)

So that,

f(a) – Ag(a) = f(b) – A.g(b) , which gives

A = …………………………….(2)

Now, h(x) is continuous in [a,b] as RHS of eq. (1) is continuous in [a,b] and h(x) is diff. In (a,b) as RHS of eq. (1) is diff. In (a,b)

Also,

h(a) = h(b)

Therefore, all the conditions of Rolle’s theorem are satisfied then there exists a Value x = cϵ (a, b)

So that h’(c) = 0

Differentiate eq.(1) w.r.t.x, we get

h’(x) = f’(x) – A.g’(x)

At x = c

h’(c) = f’(c) – A.g’(c)

0 = f’(c) – A.g’(c)

A =

So that , we get

where a<c<b

Hence the Cauchy’s mean value theorem is proved.

Example 1: Verify Cauchy’s mean value theorem for the function f(x) = sin x and g(x) = cosx in [ 0, π/2]

Solution: It is given the,

f(x) = sin x and g(x) = cos x

Now,

f’(x) = cos x and g’(x) = - sin x

We know that both the functions are continuous in [ 0 , π/2] and differentiable in ( 0 , π/2 )

Also, g’(x) = -sin x ≠ 0 for all x ϵ( 0 , π/2 )

By Cauchy’s mean value theorem, we get

for some c: 0< c <

That means

which gives,

Cot c = 1

C =

Now we see that lies between 0 and

Example 2: Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]

Solution: We are given, f(x) = x⁴ and g(x) = x

Derivative of these functions,

f’(x) = 4x³ and g’(x) = 2x

Put these values in Cauchy’s formula, we get

2c² =

c² =

c =

Now put the values of a = 1 and b = 2 ,we get

c = = = (approx)

Hence the Cauchy’s theorem is verified.

Taylor’s Theorem-

If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-

+ …….+ + ……..

Which is called Taylor’s theorem.

If we put x = a, we get-

+ …….+ + …….. (1)

Maclaurin’s Theorem-

If we put a = 0 and h = x then equation(1) becomes-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)

We get-

+ …….

Example 1: Express the polynomial in powers of (x-2).

Solution: Here we have,

f(x) =

Differentiating the function w.r.t.x-

f’(x) =

f’’(x) = 12x + 14

f’’’(x) = 12

f’’’’(x)=0

Now using Taylor’s theorem-

+ ……. (1)

Here we have, a = 2,

Put x = 2 in the derivatives of f(x), we get-

f(2) =

f’(2) =

f’’(2) = 12(2)+14 = 38

f’’’(2) = 12 and f’’’’(2) = 0

Now put a = 2 and substitute the above values in equation (1), we get-

Example 2: Expand sin x in powers of

Solution: Let f(x) = sin x

Then,

=

By using Taylor’s theorem-

+ ……. (1)

Here f(x) = sin x and a = π/2

f’(x) = cosx, f’’(x) = - sin x, f’’’(x) = - cos x and so on.

Putting x = π/2, we get

f(x) = sin x = = 1

f’(x) = cos x = = 0

f’’(x) = -sin x = = -1

f’’’(x) = -cos x = = 0

From equation (1) put a = and substitute these values, we get-

+ …….

= ………………………..

Textbooks:

1. Erwin Kreyszig, Advanced Engineering Mathematics, 10/e, John Wiley & Sons, 2011.

2. B. S. Grewal, Higher Engineering Mathematics, 44/e, Khanna Publishers, 2017.

References:

1. R. K. Jain and S. R. K. Iyengar, Advanced Engineering Mathematics, 3/e, Alpha Science

International Ltd., 2002.

2. George B. Thomas, Maurice D. Weir and Joel Hass, Thomas Calculus, 13/e, Pearson

Publishers, 2013.

3. Glyn James, Advanced Modern Engineering Mathematics, 4/e, Pearson publishers, 201.