Informative line

### Lmvt And Newton Raphson Method

Learn example of mean value theorem problems for derivatives and practice Newton Raphson Method in calculus, Use Newton's Method to find approximate value.

# The Mean Value Theorem or LaGrange's Mean Value Theorem (LMVT)

Let $$f$$ be a function defined in [a, b] such that

1. $$f$$is continuous in closed interval [a, b ]
2. $$f$$ is differentiable in open interval (a, b)

then, there exists at least one number $$c\in(a,b)$$ such that

$$f'(c)=\dfrac {f(b)-f(a)}{b-a}$$ or $$(b-a)\;f'(c)=f(b)-f(a)$$

• LMVT essentially says that the slope of line joining any two points on the graph of a function is equal to slope of a tangent of at least one point in between.

#### Which of the following function doesn't satisfy the hypotheses of LMVT in the indicated intervals?

A $$f(x) = 4x^3+7x^2-x+1$$ in [–2, 5 ]

B $$f(x) = sinx+2$$ in [5, 7 ]

C $$f(x) = tanx$$ in $$[0, \pi ]$$

D $$f(x)=\dfrac {1}{x}$$ in [5, 9 ]

×

For option (A),

$$f$$ is a polynomial function, it is continuous and differentiable every where.

$$\therefore$$ LMVT is applicable in any interval and hence in [–2, 5].

For option (B),

$$f$$ is continuous and differentiable for all values of $$x$$ having LMVT applicable in [5, 7].

For option (C),

$$f$$ is non differentiable and discontinuous at $$x=\dfrac {\pi}{2}$$

$$\therefore$$ discontinuous in interval $$[0, \pi]$$.

$$\therefore$$ LMVT is not applicable.

For option (D),

$$f$$ is discontinuous at  $$x=0$$, but is continuous and differentiable in [5, 9]

$$\therefore$$ LMVT is applicable.

### Which of the following function doesn't satisfy the hypotheses of LMVT in the indicated intervals?

A

$$f(x) = 4x^3+7x^2-x+1$$ in [–2, 5 ]

.

B

$$f(x) = sinx+2$$ in [5, 7 ]

C

$$f(x) = tanx$$ in $$[0, \pi ]$$

D

$$f(x)=\dfrac {1}{x}$$ in [5, 9 ]

Option C is Correct

# Finding the Value of Number 'c' that satisfies LMVT

• Given a function f(x) which satisfy the hypothesis of LMVT in [a, b], solve, $$f ' (x) = \dfrac {f(b) - f(a)}{b-a}$$ in [a, b].
• The solution to this equation (they can be more than one) will be called the number c(or c's)

Step 1 : Find $$f'(x)$$ i.e. derivative of $$f(x)$$.

Step 2 : Put$$f ' (x) = \dfrac {f(b) - f(a)}{b-a}$$

Step 3 : Find roots of the above equation, these will be the values of 'c'

#### Find the number 'c' that satisfies the conclusion of LMVT for following function:  $$f(x) = x^2 -2x + 4$$ in [1, 5]

A c = –1

B c = 3

C c = 4/7

D c = 5/2

×

Step 1 : Find $$f'(x)$$ i.e. derivative of $$f(x)$$.

Step 2 : Put$$f ' (x) = \dfrac {f(b) - f(a)}{b-a}$$

Step 3 : Find roots of the above equation, these will be the values of 'c'

$$f(x) = x^2 -2x + 4$$ in [1, 5]

To find c, solve

$$f ' (x) = \dfrac {f(5) - f(1)}{5-1}$$

$$\Rightarrow 2x-2=\dfrac {(25-10+4)-(1-2+4)}{4}$$

$$\Rightarrow 2x-2=\dfrac {16}{4}$$

$$\Rightarrow 2x-2=4$$

$$\Rightarrow x=3$$

Since, $$3\in (1, 5)$$

$$\Rightarrow c = 3$$

### Find the number 'c' that satisfies the conclusion of LMVT for following function:  $$f(x) = x^2 -2x + 4$$ in [1, 5]

A

c = –1

.

B

c = 3

C

c = 4/7

D

c = 5/2

Option B is Correct

#### Find the number 'c' that satisfies the conclusion of LMVT for following function  $$f(x) = (x-1)(x-2)(x-3)$$ in [0, 4]

A c =$$2\pm\dfrac {2}{\sqrt 3}$$

B c = $$4\pm\dfrac {2}{\sqrt 3}$$

C c = $$1\pm\sqrt 3$$

D c = $$\pm5$$

×

$$f(x) = (x-1)(x-2)(x-3)$$ in [0, 4]

To find c, solve

$$f ' (x) = \dfrac {f(4) - f(0)}{4-0}$$

$$\Rightarrow f'(x)=\dfrac {(3×2×1)-(-1×-2×-3)}{4-0}$$

$$\Rightarrow f'(x)=\dfrac {12}{4}$$

$$\Rightarrow f'(x)=3$$

Now, $$f '(x) = (x-1)(x-2)+(x-1)(x-3)+(x-2)(x-3)$$

(Apply product rule for three functions)

$$\Rightarrow\;f '(x) = 3x^2-12x+11$$

$$\therefore$$ $$3x^2-12x+11=3$$

$$\Rightarrow 3x^2-12x+8=0$$

$$\Rightarrow x=\dfrac {12\pm\sqrt {144-4×3×8}}{2×3}$$

$$\Rightarrow x= {2\pm }\dfrac{2}{\sqrt 3}$$

$${2+}\dfrac{2}{\sqrt 3}\in(0,4)$$ and $${2-}\dfrac{ 2}{\sqrt 3}\in(0,4)$$

Hence, two values of $$c= {2\pm}\dfrac{ 2}{\sqrt 3}$$

### Find the number 'c' that satisfies the conclusion of LMVT for following function  $$f(x) = (x-1)(x-2)(x-3)$$ in [0, 4]

A

c =$$2\pm\dfrac {2}{\sqrt 3}$$

.

B

c = $$4\pm\dfrac {2}{\sqrt 3}$$

C

c = $$1\pm\sqrt 3$$

D

c = $$\pm5$$

Option A is Correct

# Finding an Upper or Lower Bound of a Function Using LMVT

By LMVT,

$$f'(c)= \dfrac {f(b)-f(a)}{b-a}$$

for some $$c\in(a,b)$$

• Now if there is a bound given on $$f(x)$$ for all  $$x$$, and value of one of  $$f(a)$$ and $$f(b)$$, the value of other will be in a bound.
• Use $$f(b) = f'(c) (b - a) + f(a)$$

#### Let  $$f(2) =7$$  and  $$f'(x) \geq 5$$  for  $$2 \leq x\leq 8$$, what is the least possible value of $$f(8)$$?

A 35

B –18

C 37

D 49

×

By LMVT

$$f'(c)= \dfrac {f(b)-f(a)}{b-a}$$,               for some $$c\in(a,b)$$

$$\Rightarrow$$$$f'(c)= \dfrac {f(8)-f(2)}{8-2}$$, for some $$c\in(2,8)$$

$$\Rightarrow$$ $$f'(c)= \dfrac {f(8)-7}{6}$$, for some $$c\in(2,8)$$

$$\Rightarrow$$$$f(8)=6f'(c)+7$$

for some $$c\in(2,8)$$

Now,

$$\Rightarrow f ' (x) \geq 5\\ \Rightarrow\; f ' (c) \geq 5 \\\Rightarrow\; 6 f' (c) \geq 30$$

$$\therefore \; f(8)\geq30+7$$

$$\Rightarrow f(8) \geq 37$$

$$\therefore$$ least possible value of $$f(8)$$ is 37

### Let  $$f(2) =7$$  and  $$f'(x) \geq 5$$  for  $$2 \leq x\leq 8$$, what is the least possible value of $$f(8)$$?

A

35

.

B

–18

C

37

D

49

Option C is Correct

# Newton - Raphson Method

• It is a method used to approximate the roots of an equation. There are some equations whose exact roots are very hard to find, we apply this method to find an approximate root.
• Consider the graph of a function $$'f'$$.
• $$f(x) = 0$$ has roots $$\alpha$$ which we don't know.

• Take any value of $$x$$ say  $$x=x_1$$, draw tangent to the graph at $$(x, f(x))$$, let it cut the $$x$$ axis at $$x=x_2$$. Now again drawn a tangent at the point $$(x_2, f(x_2))$$. Let this intersect $$x$$ axis at $$x_3$$, observe that these values are approaching $$\alpha$$ as we do these steps again and again,
• $$\lim\limits_{x\rightarrow\infty}\;x_n=\alpha$$
• Consider the graph of $$f$$

• Tangent at $$(x_1, f(x_1))$$ has the equation

$$y-f(x_1)=f'(x_1)(x-x_1)\;...(i)$$

this will intersect $$x$$ axis at $$x_2$$  (Put $$y=0$$)

$$-f(x_1)=f'(x_1)(x_2-x_1)$$

$$\Rightarrow\;x_2=x_1-\dfrac {f(x_1)}{f'(x_1)} \Rightarrow\;x_3=x_2-\dfrac {f(x_2)}{f'(x_2)}$$

$$\Rightarrow\;x_{n+1}=x_n-\dfrac {f(x_n)}{f'(x_n)}$$

We can find better approximate by increasing values of  $$n$$.

• The choice of $$x$$, can be made by drawing a rough sketch of $$'f'$$ if possible, otherwise it can be any value.

• Suppose we need to find the value of $$\sqrt[4] {3}$$, we know that it is the root of equation $$x^4-3=0$$. Now use Newton-Raphson Method.

#### Starting with $$x_1=2$$, find third approximate root  $$x_3$$ of the equation $$3x^3+2x-5=0$$

A $$x_3=1.75$$

B $$x_3=8.34$$

C $$x_3=2.76$$

D $$x_3=1.09$$

×

Apply Newton's Method,

$$\Rightarrow\;x_{n+1}=x_n-\dfrac {f(x_n)}{f'(x_n)}$$

$$f(x)=3x^3+2x-5$$

$$\Rightarrow f'(x)=9x^2+2$$

$$\therefore\;x_2=x_1-\dfrac {f(x_1)}{f'(x_1)}$$ $$\Rightarrow\;x_2=2-\dfrac {f(2)}{f'(2)}$$

$$\Rightarrow\;x_2=2-\dfrac {23}{38}=2-.6052=1.3948$$

$$\therefore\;x_3=x_2-\dfrac {f(x_2)}{f'(x_2)}$$

$$\therefore\;x_3=1.3948-\dfrac {f(1.3948)}{f'(1.3948)}$$

$$\Rightarrow\;x_3=1.3948-\dfrac {5.9302}{19.5092}=1.3948-.3039=1.0909$$

### Starting with $$x_1=2$$, find third approximate root  $$x_3$$ of the equation $$3x^3+2x-5=0$$

A

$$x_3=1.75$$

.

B

$$x_3=8.34$$

C

$$x_3=2.76$$

D

$$x_3=1.09$$

Option D is Correct

#### Use Newton's Method to find approximate value of $$\sqrt [5]{10}$$. Find fourth approximate  $$x_4$$ and start with $$x_1=1$$

A $$x_4=1.893$$

B $$x_4=.962$$

C $$x_4=5.372$$

D $$x_4=1.125$$

×

Apply Newton's Method,

$$\Rightarrow\;x_{n+1}=x_n-\dfrac {f(x_n)}{f'(x_n)}$$

To find the value of $$\sqrt [5]{10}$$

Let $$x=10^{1/5}$$ , solve the equation

$$x^5-10=0$$

$$f(x)=x^5-10\\\Rightarrow f(x)=5x^4$$

$$\therefore\;x_2=x_1-\dfrac {f(x_1)}{f'(x_1)}$$ (Choose $$x_1=1$$)

$$\Rightarrow\;x_2=1-\dfrac {f(1)}{f'(1)}\\ =1-\dfrac {(-9)}{5}\\=\dfrac {14}{5}\\=2.8$$

$$\therefore\;x_3=x_2-\dfrac {f(x_2)}{f'(x_2)}$$

$$=2.8-\dfrac {f(2.8)}{f'(2.8)}$$

$$=2.8-\dfrac {(162.1037)}{307.328}$$

$$=2.8-0.5275\\=2.2725$$

$$\therefore\;x_4=x_3-\dfrac {f(x_3)}{f'(x_3)}$$

$$=2.2725-\dfrac {50.6065}{133.3477}$$

$$=2.2725-0.3795\\=1.893$$

### Use Newton's Method to find approximate value of $$\sqrt [5]{10}$$. Find fourth approximate  $$x_4$$ and start with $$x_1=1$$

A

$$x_4=1.893$$

.

B

$$x_4=.962$$

C

$$x_4=5.372$$

D

$$x_4=1.125$$

Option A is Correct

# Cases in which Newton's Method Fail to Approximate

• Sometimes Newton's method fails to find the approximate root of an equation. It might even happen that the approximation $$n_r$$ for some r, lies outside the domain of the function.
• Observe the following figure.

• Tangent at $$(x_1, f(x_1))$$ cuts $$x$$ axis at $$x_2$$ (actual root is $$\alpha$$)
• Now tangent at $$x_2$$ cuts $$x$$ axis at $$x_3$$ and $$x_3$$ is not in the domain of $$f$$ i.e. $$f(x_3)$$ is not defined.
• Change of choice of $$x_1$$ can lead to a good approximation.

#### Newton's method fails in finding the root of equation $$x^3-12x+2=0$$, if initial approximation $$x_1=2$$. Why?

A $$x^3-12x+2=y$$ has a horizontal tangent at x = 2

B $$x^3-12x+2=y$$ has a vertical tangent at x = 2

C $$x^3-12x+2=y$$ has a discontinuous at x = 2

D $$x^3-12x+2=y$$ has a non differentiable at x = 2

×

If $$x _1= 2$$, then $$x _ 2=x_1-\dfrac {f(x_1)}{f'(x_1)}$$

Now,

$$f'(x_1)=0$$

So, $$x_2$$ is undefined.

$$\therefore$$  We cannot prove further in general if any initial approximation is the point of horizontal tangency, we cannot proceed by Newton's Method.

### Newton's method fails in finding the root of equation $$x^3-12x+2=0$$, if initial approximation $$x_1=2$$. Why?

A

$$x^3-12x+2=y$$ has a horizontal tangent at x = 2

.

B

$$x^3-12x+2=y$$ has a vertical tangent at x = 2

C

$$x^3-12x+2=y$$ has a discontinuous at x = 2

D

$$x^3-12x+2=y$$ has a non differentiable at x = 2

Option A is Correct

#### Let $$f(x) = m\,x +5$$  if $$x \leq 1$$     $$= 4x^2 +x+n$$  if  $$x>1$$ Find the values of m and n for which LMVT can be applied to $$f$$ in the interval $$[0,2]$$.

A $$m= 9, \,n=19$$

B $$m= 90, \,n=9$$

C $$m= 19, \,n=9$$

D $$m= 9, \,n=9$$

×

If $$'f'$$ is a function defined in $$[a,b]$$ then LMVT is applicable if

1. It is continuous in  $$[a,b]$$.

2. It is differentiation in (a,b).

In this case,

$$f(x) = mx+5$$   if  $$x \leq1$$

$$= 4x^2+x+n$$    if $$x\geq 1$$

This  $$f$$ is continuous and differential for all values of $$x$$, except perhaps at $$x = 1$$ .

(It is a polynomial everywhere)

For continuity at  $$x = 1 \to$$  $$R.H.L. = L.H.L. = f(1)$$

$$R.H.L. = \lim\limits_{x\to1^+} f(x) = \lim\limits_{x\to 1^+} 4x^2+x+n = 5+n$$

$$L.H.L. = \lim\limits_{x\to1^-} f(x) = \lim\limits_{x\to 1^-} (mx+5) = m+5$$

$$\Rightarrow f(1) = m+5$$

$$\therefore 5+n = m+5\\ \Rightarrow n=m$$

For differentiable  at $$x =1, \,\,R.H.D. = L.H.D.$$

$$R.H.D. = 8+1=9$$

$$L.H.D. = m$$

$$\therefore 9=m$$

$$\therefore n=m=9$$

### Let $$f(x) = m\,x +5$$  if $$x \leq 1$$     $$= 4x^2 +x+n$$  if  $$x>1$$ Find the values of m and n for which LMVT can be applied to $$f$$ in the interval $$[0,2]$$.

A

$$m= 9, \,n=19$$

.

B

$$m= 90, \,n=9$$

C

$$m= 19, \,n=9$$

D

$$m= 9, \,n=9$$

Option D is Correct