# Newton's Method Worksheet

Newton's Method Worksheet
• Page 1
1.
Using Newton's approximation method, find the zero of the function $x$2 - 55.
 a. ± 7.4161 b. ± 7.4928 c. ± 7.2137 d. ± 7.3973

#### Solution:

f(x) = x2 - 55

f ′(x) = 2x
[Differentiate.]

From the graph, the zeros of the function are in between - 7 & - 8 and + 7 & + 8.

The Newton's approximating sequence is generated by the formula: xn+1 = xn - f(xn)f(xn)

Let x1 = + 7
[Starting point.]

x2 = 7 - (7)2 - 552(7)= 7.4285
[Substitute n = 1 in the formula.]

Similarly, x3 = 7.4162; x4 = 7.4161; x5 = 7.4161

Since the values of x4 & x5 are very close to each other, the approximate root is + 7.4161.

Let x1 = - 7
[Starting point.]

x2 = -7 - (-7)2 - 552(-7) = - 7.4285
[Substitute n = 1 in the formula.]

Similarly, x3 = - 7.4162; x4 = - 7.4161; x5 = - 7.4161

Since the values of x4 & x5 are very close to each other, hence the approximate root is - 7.4161.

Hence, the approximate roots of f(x) are 7.4161 & - 7.4161.

2.
Use Newton's method for approximating the zero(s) of the function $f$($x$) = 2$x$2 - 2$x$ - 4.
 a. 2.1000 b. - 2.0000 c. - 1.0000 d. Both - 2.0000 & - 1.0000

#### Solution:

f(x) = 2x2 - 2x - 4

f ′(x) = 4x - 2
[Differentiate.]

From the graph, the zeros of the function are - 1 & 2.

Though we are getting the roots from the graph directly, Let us check the zeros of the function by using " NewtonÃ¢â‚¬â„¢s method".

The Newton's approximating sequence is generated by the formula: xn+1 = xn - f(xn)f (xn).

Let x1 = 1.5
[Starting point.]

x2 = 1.5 - 2(1.5)2 - 2(1.5)  -  44(1.5) - 2 = 2.125.

Similarly, x3 = 2.0048; x4 = 2.00001; x5 = 2.0000; x6 = 2.0000
[Substitute n = 1 in the formula.]

Since, the values of x5 & x6 are very close to each other, the approximate solution of x is 2.0000.

Let x1 = - 0.5
[Starting point.]

x2 = -0.5 - 2(-0.5)2 - 2(-0.5)  -  44(-0.5) - 2 = - 1.125;
[Substitute n = 1 in the formula.]

Similarly, x3 = - 1.0048; x4 = - 1.00001; x5 = - 1.0000; x6 = - 1.0000.

Since, the values of x5 & x6 are very close to each other, the other approximate solution of x is - 1.0000.

Hence, the approximate zeros of f(x) are 2.0000 & - 1.0000.

3.
Use Newton's method for approximating the zeros of the function $f$($x$) = 3$x$2 - 28.
 a. ± 3.113 b. ± 3.055 c. ± 9.235 d. ± 9.332

#### Solution:

f(x) = 3x2 - 28

f ′(x) = 6x
[Differentiate.]

From the graph, the zeros of the function lie between - 3 & - 4 and 3 & 4.

The Newton's approximating sequence is generated by the formula: xn+1 = xn - f(xn)f (xn)

Let, x1 = - 3
[Starting point.]

x2 = -3 - 3(-3)2 - 286(-3) = - 3.0555
[Substitute n = 1 in the formula.]

Similarly x3 = - 3.055; x4 = - 3.055

Since the values of x3 & x4 are very close to each other, hence the approximate root is - 3.055.

Let x1 = 3
[Starting point.]

x2 = 3 - 3(3)2 - 286(3) = 3.055
[Substitute n = 1 in the formula.]

Similarly x3 = 3.055; x4 = 3.055

Since the values of x3 & x4 are very close to each other, the approximate root is 3.055.

Hence, the approximate roots of f(x) are - 3.055 & 3.055.

4.
Use Newton's method for approximating the roots of the function $f$($x$) = $x$4 + 3$x$3 + 5.
 a. -1.00438 & -3.6611 b. -1.2234 & -3.7861 c. -1.49074 & -2.7629 d. -1.36827 & -2.7629

#### Solution:

f(x) = x4 + 3x3 + 5

f′(x) = 4x3 + 9x2
[Differentiate.]

From the graph, the zeros of function are in between - 1 & - 2 and - 2 & - 3.

The Newton's approximating sequence is generated by the formula: xn+1 = xn - f(xn)f (xn).

Let x1 = - 1.5
[Starting point.]

x2 = -1.5 - (-1.5)4 + 3(-1.5)3 + 54(-1.5)3 + 9(-1.5)2 = - 1.49074
[Substitute n = 1 in the formula.]

Similarly, x3 = - 1.49074; x4 = - 1.49074

Since the values of x3 & x4 are very close to each other, the approximate root is - 1.49074.

Let x1 = - 2.5
[Starting point.]

x2 = -2.5 - (-2.5)4 + 3(-2.5)3 + 54(-2.5)3 + 9(-2.5)2 = - 2.95
[Substitute n = 1 in the formula.]

Similarly, x3 = - 2.7974; x4 = - 2.7644; x5 = - 2.7629; x6 = - 2.7629

Since the values of x5 & x6 are very close to each other, the approximate root is - 2.7629.

Hence, the approximate roots of f(x) are - 1.49074 & - 2.7629.

5.
Use Newton's method for approximating the root(s) of the function $f$($x$) = 12$x$2 - 5$x$ - 19.
 a. -1.06710 and 1.48376 b. 1.48376 c. - 1.06710 d. 1.06710

#### Solution:

f(x) = 12x2 - 5x - 19

f ′(x) = 24x - 5
[Differentiate.]

From the graph, the zeroÃ¢â‚¬â„¢s of function are in between 0 & - 1 and 1 & 2.

The Newton's approximating sequence is generated by the formula: xn+1 = xn - f(xn)f (xn).

Let x1 = - 1
[Starting point.]

x2 = -1 - 12(-1)2 - 5(-1) - 1924(-1) - 5 = - 1.0689
[Substitute n = 1 in the formula.]

Similarly x3 = - 1.06710; x4 = - 1.06710

Since, the values of x3 & x4 are very close to each other, the approximate root is - 1.06710.

Let x1 = 1.2
[Starting point.]

x2 = 1.2 - 12(1.2)2 - 5(1.2) - 1924(1.2) - 5 = 1.52436
[Substitute n = 1 in the formula.]

Similarly, x3 = 1.48439; x4 = 1.48376; x5 = 1.48376

Since, the values of x4 & x5 are very close to each other, the approximate root is 1.48376.

Hence, the approximate roots of f(x) are - 1.06710 & 1.48376.

6.
Identify the Newton's iterative formula that helps to approximate the 5th root of the positive number $k$.
 a. ${x}_{n+1}=4{x}_{n}$ b. ${x}_{n+1}=-\frac{1}{5}\left(4{x}_{n}+\frac{k}{{{x}_{n}}^{4}}\right)$ c. ${x}_{n+1}=4{x}_{n}+\frac{k}{{{x}_{n}}^{4}}$ d. ${x}_{n+1}=\frac{1}{5}\left(4{x}_{n}+\frac{k}{{{x}_{n}}^{4}}\right)$

#### Solution:

Let x be the 5th root of the positive number k.

So, f (x) = x5 - k = 0

f ′ (x) = 5x4
[Evaluate f ′ (x).]

xn+1=xn-f(xn)f(xn)
[Write the Newton's iterative formula.]

= xn-xn5-k5xn4 = xn-xn5+k5xn4

= 1 / 5(4xn+kxn4)

So, xn+1=15(4xn+kxn4) is the Newton's iterative formula to approximate the 5th root of the positive number k.

7.
Find the roots of $f$($x$) = 2$x$2 - 17.
 a. ± 2.7865 b. ± 2.5569 c. ± 2.9154 d. ± 2.6781

#### Solution:

f (x) = 2x2 - 17

f ′(x) = 4x
[Differentiate.]

From the graph, the zeros of function are in between - 3 & - 2 and 2 & 3.

The Newton's approximating sequence is generated by the formula, xn + 1 = xn - f(xn)f(xn).

Let x1 = - 2
[Starting point.]

x2 = -2 - 2(-2)2 - 174(-2) = - 3.125
[Substitute n = 1 in the formula.]

Similarly x3 = - 2.9225; x4 = - 2.9154; x5 = - 2.9154

Since the values of x4 & x5 are very close to each other, the approximate root is - 2.9154.

Let x1 = 2
[Starting point.]

x2 = 2 - 2(2)2 - 174(2) = 3.125
[Substitute n = 1 in the formula.]

Similarly x3 = 2.9225; x4 = 2.9154; x5 = 2.9154

Since the values of x4 & x5 are very close to each other, the approximate root is 2.9154.

Hence, the approximate roots of f (x) are - 2.9154 & 2.9154.