Unit 2 | unit 2 mean value theorems

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Unit 2

Mean value theorems

2.1 Rolle’s theorem

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

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.

2.2 Lagrange’s mean value theorem

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.

2.3 Cauchy’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.

2.4 Taylor’s and Maclaurin theorems with remainders (without proof)

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-

+ …….+ + ……..