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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.

Illustration Questions

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'

Illustration Questions

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

Illustration Questions

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 \( f(b) = f'(c) (b - a) + f(a)\)

Illustration Questions

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.

Illustration Questions

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

Illustration Questions

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.

Illustration Questions

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.

image

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

image

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

Illustration Questions

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

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