Use Newton's method with initial approximation

x1 = −2
to find x2, the second approximation to the root of the equation
x^3 + x + 1 = 0.

*I got -1.307692 for my answer but it said that was wrong, so then I tried rounding it to two decimal places and got -1.31 and that still didn't work and then i tried -1.30 and it still didn't work

Visit

http://keisan.casio.com/exec/system/1244946907

and type in your functions. You will see that the root is -0.68233

You can also see this at wolframalpha.com

When you try to solve a problem and don't get a satisfactory answer, it would help if you showed your work. Then we could see where you went astray.

-0.68233 is still wrong. I had tried that one earlier too

To use Newton's method, we start with an initial approximation and refine it using the formula:

x2 = x1 - (f(x1)) / (f'(x1))

Let's apply this formula to find x2, the second approximation to the root of the equation x^3 + x + 1 = 0.

First, we need to find the derivative of f(x) = x^3 + x + 1. The derivative is:

f'(x) = 3x^2 + 1

Now, let's substitute the values into the formula:

x2 = x1 - (f(x1)) / (f'(x1))
= -2 - ((-2)^3 + (-2) + 1) / (3(-2)^2 + 1)
= -2 - (-8 - 2 + 1) / (3(4) + 1)
= -2 - (-9) / (12 + 1)
= -2 + 9 / 13
= -2 + 0.6923076923076923
= -1.3076923076923075

Therefore, the second approximation to the root of the equation x^3 + x + 1 = 0 using Newton's method with an initial approximation of x1 = -2 is approximately -1.31 (rounded to two decimal places). It seems like you calculated it correctly, so there might be a mistake in the input or the expected answer. Please double-check the question or consult with your instructor if there is any confusion.

To use Newton's method to find the second approximation, x2, of the root of the equation x^3 + x + 1 = 0, follow these steps:

Step 1: Write down the equation and its derivative.
- Equation: f(x) = x^3 + x + 1
- Derivative: f'(x) = 3x^2 + 1

Step 2: Start with the initial approximation, x1 = -2, and calculate the next approximation using the formula:
- x2 = x1 - f(x1) / f'(x1)

Step 3: Substitute the value of x1 into the equation and its derivative to calculate f(x1) and f'(x1).

f(x1) = (-2)^3 + (-2) + 1 = -8 - 2 + 1 = -9
f'(x1) = 3(-2)^2 + 1 = 12 + 1 = 13

Step 4: Calculate x2 using the formula:
- x2 = -2 - (-9) / 13 = -2 + 9 / 13

Step 5: Simplify the expression:
- x2 = (-2 * 13 + 9) / 13 = (-26 + 9) / 13 = -17 / 13

Step 6: Round the value to the desired number of decimal places.
- x2 ≈ -1.30769231 (unrounded)
- x2 ≈ -1.31 (rounded to two decimal places)

It seems that you performed the calculations correctly. Since the answer you got (-1.307692) isn't being accepted, it's possible that there is a rounding error or the answer is expected to be rounded to a different number of decimal places.

If you've tried rounding to two decimal places (-1.31) or one decimal place (-1.3) and it's still not accepted, consider checking the instructions or requirements provided by your course or instructor to ensure you're formatting the answer correctly.