Calculus Examples

x^{3} - 7 = 0

$x^{3} - 7 = 0$ ,

a = 2

$a = 2$

Step 1

Find the derivative of

f (x) = x^{3} - 7

$f (x) = x^{3} - 7$ for use in Newton's method.

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Step 1.1

By the Sum Rule, the derivative of

x^{3} - 7

$x^{3} - 7$ with respect to

x

$x$ is

\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 7]

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 7]$ .

\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 7]

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 7]$

Step 1.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 3

$n = 3$ .

3 x^{2} + \frac{d}{d x} [- 7]

$3 x^{2} + \frac{d}{d x} [- 7]$

Step 1.3

Since

- 7

$- 7$ is constant with respect to

x

$x$ , the derivative of

- 7

$- 7$ with respect to

x

$x$ is

0

$0$ .

3 x^{2} + 0

$3 x^{2} + 0$

Step 1.4

Add

3 x^{2}

$3 x^{2}$ and

0

$0$ .

3 x^{2}

$3 x^{2}$

3 x^{2}

$3 x^{2}$

Step 2

Set up the formula to find the

x_{2}

$x_{2}$ approximation.

$x_{2} = x_{1} - \frac{f (x_{1})}{f' (x_{1})}$

Step 3

Substitute the value of

$x_{1}$ into the next Newton's Method approximation.

$x_{2} = 2 - \frac{{(2)}^{3} - 7}{3 {(2)}^{2}}$

Step 4

Simplify the right side of the equation to find

$x_{2}$ .

$x_{2} = 1.91\overline{6}$

Step 5

Set up the formula to find the

$x_{3}$ approximation.

$x_{3} = x_{2} - \frac{f (x_{2})}{f' (x_{2})}$

Step 6

Substitute the value of

$x_{2}$ into the next Newton's Method approximation.

$x_{3} = 1.91\overline{6} - \frac{{(1.91\overline{6})}^{3} - 7}{3 {(1.91\overline{6})}^{2}}$

Step 7

Simplify the right side of the equation to find

$x_{3}$ .

$x_{3} = 1.91293845$

Step 8

Set up the formula to find the

$x_{4}$ approximation.

$x_{4} = x_{3} - \frac{f (x_{3})}{f' (x_{3})}$

Step 9

Substitute the value of

$x_{3}$ into the next Newton's Method approximation.

$x_{4} = 1.91293845 - \frac{{(1.91293845)}^{3} - 7}{3 {(1.91293845)}^{2}}$

Step 10

Simplify the right side of the equation to find

$x_{4}$ .

$x_{4} = 1.91293118$

Step 11

Set up the formula to find the

$x_{5}$ approximation.

$x_{5} = x_{4} - \frac{f (x_{4})}{f' (x_{4})}$

Step 12

Substitute the value of

$x_{4}$ into the next Newton's Method approximation.

$x_{5} = 1.91293118 - \frac{{(1.91293118)}^{3} - 7}{3 {(1.91293118)}^{2}}$

Step 13

Simplify the right side of the equation to find

$x_{5}$ .

$x_{5} = 1.91293118$

Step 14

Set up the formula to find the

$x_{6}$ approximation.

$x_{6} = x_{5} - \frac{f (x_{5})}{f' (x_{5})}$

Step 15

Substitute the value of

$x_{5}$ into the next Newton's Method approximation.

$x_{6} = 1.91293118 - \frac{{(1.91293118)}^{3} - 7}{3 {(1.91293118)}^{2}}$

Step 16

Simplify the right side of the equation to find

$x_{6}$ .

$x_{6} = 1.91293118$

Step 17

Since the

$6 t h$ and

$5 t h$ approximations are equal to

$6$ decimal places,

$1.91293118$ is the approximation of the root.

$1.91293118$