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

Let \(f\) be a function defined in [a, b] such that

- \(f\)is continuous in closed interval [a, b ]
- \(f\) 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.

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 ]

- 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'

A c = –1

B c = 3

C c = 4/7

D c = 5/2

A c =\(2\pm\dfrac {2}{\sqrt 3}\)

B c = \(4\pm\dfrac {2}{\sqrt 3}\)

C c = \(1\pm\sqrt 3\)

D c = \(\pm5\)

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 \( f(b) = f'(c) (b - a) + f(a)\)

- 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.

A \(x_3=1.75\)

B \(x_3=8.34\)

C \(x_3=2.76\)

D \(x_3=1.09\)

A \(x_4=1.893\)

B \(x_4=.962\)

C \(x_4=5.372\)

D \(x_4=1.125\)

- 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.

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

A \(m= 9, \,n=19\)

B \(m= 90, \,n=9\)

C \(m= 19, \,n=9\)

D \(m= 9, \,n=9\)