please help!!!!Apply newtons method to approximate the x-values of the indicated points of intersection of the 2 graphs. continue the process until 2 successive approximations differ by less than 0.001. (Hint: Let h(x)=f(x)-g(x))
f(x)=x^6
g(x)=cos(x)
x= ??? (smaller value)
x=??? (larger value)
Thanks!!
As often is the case, the starting point can make a difference. I started at x=1.1
h(1.10000) = 1.31796 -- next x: 0.97512
h(0.97512) = 0.29866 -- next x: 0.92631
h(0.92631) = 0.03093 -- next x: 0.91998
h(0.91998) = 0.00045 -- next x: 0.91989
Starting at x=0.1, things take a bit longer to converge:
h(0.10000) = -0.99500 -- next x: 10.06065
h(10.06065) = 1036945.84148 -- next x: 8.38387
h(8.38387) = 347270.57671 -- next x: 6.98656
h(6.98656) = 116299.56298 -- next x: 5.82215
h(5.82215) = 38948.45732 -- next x: 4.85180
h(4.85180) = 13044.10696 -- next x: 4.04313
h(4.04313) = 4368.84599 -- next x: 3.36910
h(3.36910) = 1463.42243 -- next x: 2.80716
h(2.80716) = 490.27325 -- next x: 2.33854
h(2.33854) = 164.25215 -- next x: 1.94780
h(1.94780) = 54.97734 -- next x: 1.62277
h(1.62277) = 18.31398 -- next x: 1.35550
h(1.35550) = 5.98917 -- next x: 1.14486
h(1.14486) = 1.83849 -- next x: 1.00022
h(1.00022) = 0.46121 -- next x: 0.93287
h(0.93287) = 0.06355 -- next x: 0.92027
h(0.92027) = 0.00182 -- next x: 0.91989
h(0.91989) = 0.00000 -- next x: 0.91989
To apply Newton's method to approximate the x-values of the points of intersection of the two graphs, f(x) = x^6 and g(x) = cos(x), we can follow these steps:
Step 1: Define the function h(x) = f(x) - g(x).
Step 2: Find the derivative of h(x) with respect to x, denoted as h'(x).
Step 3: Choose an initial approximation for the smaller x-value and call it x_0.
Step 4: Apply the Newton's method iterative formula:
x_(n+1) = x_n - h(x_n) / h'(x_n)
where x_n is the current approximation and x_(n+1) is the next approximation.
Step 5: Repeat Step 4 until two successive approximations, let's say x_n and x_(n+1), differ by less than 0.001. In other words, continue the process until |x_(n+1) - x_n| < 0.001.
Step 6: Once you have found the smaller x-value, choose a new initial approximation for the larger x-value and repeat Steps 4 and 5 until you find the larger x-value where |x_(n+1) - x_n| < 0.001.
Let's go through the process step by step:
Step 1: Define h(x) = f(x) - g(x) = x^6 - cos(x).
Step 2: Find the derivative of h(x): h'(x) = 6x^5 + sin(x).
Step 3: Choose an initial approximation for the smaller x-value, let's say x_0 = 0.
Step 4: Apply the Newton's method iterative formula:
x_(n+1) = x_n - (x_n^6 - cos(x_n)) / (6x_n^5 + sin(x_n))
Substitute x_0 = 0 into the formula:
x_(1) = 0 - (0^6 - cos(0)) / (6(0)^5 + sin(0))
Simplifying, we get:
x_(1) = 0 - cos(0) / (sin(0))
Since cos(0) = 1 and sin(0) = 0, we have:
x_(1) = 0 - 1 / 0 = undefined
In this case, the initial approximation x_0 = 0 leads to an undefined result. We need to choose a different initial approximation.
Let's choose a new initial approximation, x_0 = 1:
x_(1) = 1 - (1^6 - cos(1)) / (6(1)^5 + sin(1))
Simplifying, we get:
x_(1) ≈ 0.907
Now we have the first approximation for the smaller x-value.
Step 5: Check if |x_(n+1) - x_n| < 0.001. If it is true, then we have found the desired approximation. Otherwise, continue with Step 4 using x_n as the new approximation.
Let's use x_1 as the new approximation and repeat Step 4:
x_(2) = x_1 - (x_1^6 - cos(x_1)) / (6x_1^5 + sin(x_1))
Repeat the process until |x_(n+1) - x_n| < 0.001.
Follow the same steps to find the larger x-value, choosing a new initial approximation.
Remember to continue the process until the difference between two successive approximations is less than 0.001 for both the smaller and larger x-values.