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3 years ago

Using Newton-Raphsf (x) = 2x3 + x2 - x + 1 accurate to within 10-4

Mathematics

2 answers:

Effectus [21]3 years ago

7 0

9514 1404 393

Answer:

-1.2338

Step-by-step explanation:

The graph shows the one real root is about -1.234. The Newton's method iterator is ...

new x = x - f(x)/f'(x) . . . . . where f'(x) is the derivative of f(x)

Many graphing calculators provide a derivative function, so the iteration equation can be evaluated more or less directly by the calculator. In the attached, we have shown a table with three iterations of the root. After the second iteration, the root is good to 4 decimal places.

x ≈ -1.2338

Sergio [31]3 years ago

4 0

Answer:

2x3+x2−x+1x

Step-by-step explanation:

Let's solve for f.

fx=2x3+x2−x+1

Step 1: Divide both sides by x.

fx=2x3+x2−x+1x

fx=2x3+x2−x1x

Answer:

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1 year ago

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vlada-n [284]

Answer:

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Step-by-step explanation:

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Hope this helps :)

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3 years ago

NEED HELP ASAP! The graph gx is a translation of the function fx=x2. The vertex of gx is located 5 units above and 7 units to th

OLga [1]

Answer:

g(x) is option 1 or g(x) = (x + 7)^2 + 5

A quadratic equation has the general form of:

y=ax² +bx + c

It can be

converted to the vertex form in order to determine the vertex of the parabola.

It has the standard form of:y =a(x+h)² + k

where hand k represents the vertex, h represent the point in the x axis and k is thepoint in the y axis. Therefore, from the details given in the problem, the equation that represents

g(x) is option 1 or g(x) = (x + 7)^2 + 5

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3 years ago

Using Newton-Raphsf (x) = 2x3 + x2 - x + 1 accurate to within 10-4​

Using Newton-Raphsf (x) = 2x3 + x2 - x + 1 accurate to within 10-4