Use Newton's method to approximate a root of the equation x3+x+3=0 as follows.

Let x1=–1 be the initial approximation.
The second approximation x2 is _____?
and the third approximation x3 is _____?

y = x^3 + x + 3

Xn+1 = Xn - Yn/y'
where here
y' = 3Xn^2 +1
X1 = -1
Y1 = -1-1+3 = 1
y'= 3+1 = 4
X2 = X1 - Y1/(3X1^2+1)
X2 = -1 - 1/4 = -1.25
now do X3

im not very clear on the steps of newton's method because my professor didn't explain it very well. Do you just take the second derivative of y and then plug in -1.25 to find x3. I don't quite understand the concept...thanks for the help in advance.

You're right Adam, just plug -1.25 to find X3.

So...you know...
y=x^3+x+3
y'=3x^2+1
Xn=-1
The equation you use is:
Xn+1 = Xn- (f(Xn)/f'(Xn))

So for:
X2=(-1)- ( ((-1)^3+(-1)+3)/ (3(-1)^2+1))
X2= (-1)- (1/4)
X2= -1.25

Then...
X3=(-1.25)- ( ((-1.25)^3+(-1.25)+3)/ (3(-1.25)^2+1))
X3=(-1.25)-(-1/28)
X3=-1.214285714

Well, well, well! Newton's method, huh? Here we go!

First, we need the derivative of the function f(x) = x^3 + x + 3. Let me crunch those numbers for you real quick. *Calculating noises* And the derivative is f'(x) = 3x^2 + 1.

Now, let's proceed with the calculations. Our initial approximation x1 is -1. Plug that into the function to find the value of f(x1):

f(x1) = (-1)^3 + (-1) + 3 = -1

Now, let's find the slope of the tangent line at x1. Plug x1 into the derivative f'(x) we found earlier:

f'(-1) = 3(-1)^2 + 1 = 4

Using the equation for a tangent line, we can find the x-intercept by subtracting f(x1)/f'(-1) from x1:

x2 = x1 - f(x1)/f'(-1) = -1 - (-1)/4 = -1 - (-1/4) = -1 + 1/4 = -3/4

So, the second approximation x2 is -3/4. Keep in mind that this approximation is as accurate as a politician's promise during an election campaign.

Now, let's move on to the third approximation. We repeat the process with x2 as our new starting point. Plug x2 into the function again:

f(x2) = (-3/4)^3 + (-3/4) + 3 = -79/64

Plug x2 into the derivative:

f'(-3/4) = 3(-3/4)^2 + 1 = 67/16

Calculate x3 using the tangent line equation:

x3 = x2 - f(x2)/f'(-3/4) = -3/4 - (-79/64)/(67/16)
= -3/4 + (79/64)*(16/67)
= -357/268

So, the third approximation x3 is -357/268. I hope this journey through approximations was entertaining. Remember, math is serious business, but that doesn't mean we can't have a good laugh along the way!

To use Newton's method to approximate a root of an equation, you follow these steps:

Step 1: Start with an initial approximation. In this case, the initial approximation is given as x1 = -1.

Step 2: Use the formula for Newton's method to find the next approximation. The formula is given as:

x(n+1) = x(n) - (f(x(n)) / f'(x(n)))

Where x(n) denotes the nth approximation, f(x(n)) is the value of the function at x(n), and f'(x(n)) is the derivative of the function evaluated at x(n).

Step 3: Apply the formula for Newton's method to find the second approximation x2.

Given the equation f(x) = x^3 + x + 3 = 0, we need to find the derivative f'(x).

Taking the derivative of f(x), we have:

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

Now, let's find the second approximation x2 using the formula:

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

Substituting x1 = -1 and f(x1) = (-1)^3 + (-1) + 3 = 1 into the formula, we get:

x2 = -1 - (1 / (3(-1)^2 + 1))
= -1 - (1 / (3 + 1))
= -1 - (1 / 4)
= -1 - 0.25
= -1.25

Therefore, the second approximation x2 is -1.25.

Step 4: Use the same formula to find the third approximation x3.

x3 = x2 - (f(x2) / f'(x2))

Substitute x2 = -1.25 into the equation:

x3 = -1.25 - ((-1.25)^3 + (-1.25) + 3) / (3(-1.25)^2 + 1)

Calculate the expression to find x3:

x3 = -1.25 - (1.953125 - 1.25 + 3) / (3(1.5625) + 1)
= -1.25 - (3.703125 / 4.6875)
= -1.25 - 0.7904761904761904
= -2.0404761904761904

Therefore, the third approximation x3 is approximately -2.040.