Newton's Method for Approximating Zeros Worksheet

Newton's Method for Approximating Zeros Worksheet
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1.
Use Newton's method to find the value of $\sqrt[3]{-6}$ to the nearest thousendth.
 a. - 2.182 b. - 1.818 c. 1.833 d. 1.818 e. - 1.833

Solution:

- 63 is a solution of the equation x3 + 6 = 0

Let, f(x) = x3 + 6

Draw the graph of the function.

From the graph, the zero of the function is in between - 2 and - 1.

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

f ′(x) = 3x2

x1 = - 2 as the starting point.
[Zero of the function lies between - 2 and - 1]

x2 = x1 - f(x1)f(x1) = - 1.833

Similarly, x3 = - 1.818, x4 = - 1.818

So, the zero of the function is - 1.818.

Therefore, - 63 = - 1.818, to the nearest thousendth.

2.
Find the approximate value of $\sqrt{\frac{3}{2}}$ to the nearest thousendth by using Newton's method.
 a. - 1.235 b. 1.235 c. 0.775 d. 0.765 e. 1.225

Solution:

32 is a solution of the equation 2x2 - 3 = 0

Let, f(x) = 2x2 - 3

Draw the graph of the function.

From the graph, the approximate value of 32 is in between 1 and 2.

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

f ′(x) = 4x

Consider x1 = 1 as the starting point.
[Zero of the function lies between 1 and 2. ]

x2 = x1 - f(x1)f(x1) = 1 - 2(1)2-34 = 1.25

Similarly, x3 = 1.235, x4 = 1.225, x5 = 1.225

One zero of the function is 1.225.

So, the approximate value of 32 is 1.225, to the nearest thousendth.

3.
Use Newton's method to find a zero of the function between - 1 and 0 to three places of decimals.
$f$($x$) = $x$3 + $x$ + 1
 a. - 0.683 b. 1.317 c. 0.683 d. - 1.317 e. - 0.686

Solution:

f(- 1) = - 1 < 0 and f(0) = 1 > 0

f(- 1) and f(0) are of opposite signs, by intermediate value theorem, there exist at least one zero of the funtion in the interval (- 1, 0)

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

f ′(x) = 3x2 + 1

x1 = - 1
[Starting point.]

x2 = x1 - f(x1)f(x1)

x2 = - 1 - (-1)3-1+13(-1)2+1 = - 0.75

Similarly, x3 = - 0.686, x4 = - 0.683, x5 = - 0.6829

Since x3 and x4 are very close to each other, the zero of x is - 0.683.

So, the zero of function f(x) is - 0.683.

4.
Use Newton's method to find a zero of the function in [2, 3] to nearest four decimals.
$f$($x$) = ${x}^{3}$ - 2$x$ - 5 = 0
 a. 2.1982 b. 2.0945 c. 2.1569 d. 2.0016 e. 2.8032

Solution:

f(2) = - 1 < 0 and f(3) = 16 > 0

f(2) and f(3) are of opposite signs, by intermediate value theorem, there exists at least one zero of the function in the interval (2, 3).

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

f ′(x) = 3x2 - 2
[Find f ′(x).]

x1 = 2
[Starting point.]

x2 = x1 - f(x1)f(x1)

= 2 - 23 - 2(2) - 53(22) - 2 = 2.1

Similarly, x3 = 2.0946 and x4 = 2.0945

So, the zero of the function in [2, 3] to nearest four decimals is 2.0945

5.
Use Newton's method to find the zero of the function between 0 and 1, to the four places of decimals.
$f$($x$) = 2$x$ - ln(2 + $x$2)
 a. 0.3465 b. 0.3716 c. - 0.3465 d. - 0.3817 e. 0.3817

Solution:

f(0) = 2(0) - ln(2 + (0)2) = - 0.3010 < 0 and f(1) = 2(1) - ln(2 + (1)2) = 1.5228 > 0
[Write the function.]

f(0) and f(1) are of opposite signs, by intermediate value theorem, there exists at least one zero of the function in the interval (0, 1).

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

f ′(x) = 2 - 2x(2+x2)

x1 = 0
[Starting point.]

x2 = x1 - f(x1)f(x1)

x2 = 0 - 2(0)-ln(2+0)4+02-0 = 0.3465

Similarly, x3 = 0.3716, x4 = 0.3817, x5 = 0.3817

Therefore, the zero of function f(x) is 0.3817, to the nearest four decimals.

6.
Use Newton's method to find the roots of the equation to four decimal places.
$x$2 - 8 = 0
 a. - 1.1716 and 1.1716 b. - 2.8334 and 2.8334 c. - 2.8284 and 2.8284 d. - 2.8284 and 1.1716 e. - 1.1716 and 2.8284

Solution:

Let f(x) = x2 - 8
[Write the function.]

Draw the graph of the function.

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

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

Let us consider - 2 as the starting point to find a root in between - 2 and - 3.

f ′(x) = 2x
[Find f ′(x).]

x1 = - 2
[Starting point.]

x2 = x1 - f(x1)f(x1)

x2 = - 2 - (-2)2-82(-2) = - 3

Similarly, x3 = - 2.8333, x4 = - 2.8284, x5 = - 2.8284

The zero of the function is - 2.8284.

Let us consider 2 as the starting point to find a root in between 2 and 3.

x1 = 2

x2 = x1 - f(x1)f(x1)

= 2 - (2)2-82(2) = 3

Similarly, x3 = 2.8333, x4 = 2.8284, x5 = 2.8284

The zero of the function is 2.8284.

Therefore, the roots of the equation are - 2.8284 and 2.8284, to four decimal places.

7.
Find the zero of the function to the nearest four decimals by using Newton's method.
$f$($x$) = 3 - $e$2$x$
 a. 0.3964 b. 0.5518 c. - 0.4507 d. 0.4481 e. 0.5493

Solution:

f(x) = 3 - e2x
[Write the function.]

Draw the graph of the function.

From the graph, the zero of the function is in between 0 and 1.

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

f ′(x) = - 2e2x
[Find f ′(x).]

x1 = 0.5
[Starting point.]

x2 = x1 - f(x1)f(x1)

= 0.5 - 3-e2(0.5)-2e2(0.5) = 0.5518

Similarly, x3 = 0.5493, x4 = 0.5493

Therefore the zero of the function f(x) is 0.5493, to the nearest four decimals.

8.
Use Newton's method to find the solution for the function in [- 1, 0] to the nearest four decimals $f$($x$) = $e$$x$ + 2$x$.
 a. 0.3333 b. - 0.3333 c. - 0.3517 d. 0.3829 e. 0.3517

Solution:

f(- 1) = - 1.6321 < 0 and f(0) = 1 > 0

f(-1) and f(0) are of opposite signs, by intermediate value theorem, there exists at least one zero of the function in the interval (- 1, 0)

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

f ′(x) = ex + 2
[Find f ′(x).]

x1 = 0
[Starting point.]

x2 = x1 - f(x1)f(x1)

= 0 - e0+2(0)e0+2 = - 1 / 3 = - 0.3333

Similarly, x3 = - 0.3517, x4 = - 0.3517

So, the solution of f(x) is - 0.3517, to the nearest 4 decimal places.

9.
Find the zero of the function in [- 1, 0] to the nearest four decimals $f$($x$) = $e$cos $x$ + 2$x$ by using the Newton's method.
 a. - 1 b. - 0.9177 c. - 1.0823 d. - 0.9179 e. - 1.1506

Solution:

f(- 1) < 0 and f(0) > 0

f(- 1) and f(0) are of the opposite signs, by intermediate value theorem, there exists at least one root of the function in the interval (- 1, 0)

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

f ′(x) = - sin x ecos x + 2
[Find f ′(x).]

x1 = - 1

x2 = x1 - f(x1)f(x1)

= - 1 - [ecos(-1)+2(-1)-sin(-1)ecos(-1)+2] = - 0.9177

Similarly, x3 = - 0.9179, x4 = - 0.9179

So, the zero of the function f(x) is - 0.9179, to the nearest four decimal places.

10.
Use Newton's method to find the zero of the function between - 1 and 0 to nearest four decimals.
$f$($x$) = 2$x$ + $\sqrt{3-{x}^{2}}$
 a. - 0.5469 b. - 1 c. - 0.7837 d. - 0.8 e. - 0.7746

Solution:

f(- 1) = - 0. 5857 < 0 and f(0) = 1.732 > 0

f( -1) and f(0) are of the opposite signs, by intermediate value theorem, there exists at least one zero of the function in the interval (- 1, 0)

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

f ′(x) = 2 - x3-x2
[Find f ′(x).]

x1 = - 1
[Starting point.]

x2 = x1 - f(x1)f(x1)

= - 1 - [2(-1)+3-(-1)22--13-(-1)2] = - 0.7837

Similarly, x3 = - 0.7746, x4 = - 0.7746

So, the zero of the function f(x) is - 0.7746, to the nearest four decimals.