Sheen1101 wrote:
I have a calculus problem that I don't understand how to do and need help......
Use Newton's method to approximate the indicated root of the equation correct to six decimal places.
The positive root of 2cos(x) = x4
Welll...
Newton's method uses the formula x
1 =
x0 - f(x)/f'(x), where x
0 is an initial guess of what the root is. You then sub x
1 into the formula to get x
2, and so on.
I would suggest using f(x) = 2cos(x) - x
4From wikipedia, which might help you:
onsider the problem of finding the positive number x with cos(x) = x3. We can rephrase that as finding the zero of f(x) = cos(x) − x3. We have f'(x) = −sin(x) − 3x2. Since cos(x) ≤ 1 for all x and x3 > 1 for x>1, we know that our zero lies between 0 and 1. We try a starting value of x0 = 0.5. (Note that a starting value of 0 will lead to an undefined result, showing the importance of using a starting point that is close to the zero.)