Answer:
Using Newton's Method, the points of intersection between f(x) = x^6 and g(x) = cos(x) are approximately (0.8241, f(0.8241)) and (2.3111, f(2.3111)), where f(x) = x^6.
To find the points of intersection of the graphs of f(x) = x^6 and g(x) = cos(x), we can solve the equation h(x) = f(x) - g(x) = x^6 - cos(x) = 0.
Explanation:
We will use Newton's Method to approximate the x-value(s) of intersection. The formula for Newton's Method is:
x_n+1 = x_n - f(x_n)/f'(x_n)
where x_n is the nth approximation of the root, f(x_n) is the function evaluated at x_n, and f'(x_n) is the derivative of the function evaluated at x_n.
Let h(x) = x^6 - cos(x), then
h'(x) = 6x^5 + sin(x)
Now we need to choose a starting value for x. By graphing the two functions, we can see that there are two points of intersection in the interval [0,1]. Let's choose x = 0.5 as our starting value.
x_0 = 0.5
x_1 = x_0 - h(x_0)/h'(x_0) = 0.5352
x_2 = x_1 - h(x_1)/h'(x_1) = 0.8656
x_3 = x_2 - h(x_2)/h'(x_2) = 0.8249
x_4 = x_3 - h(x_3)/h'(x_3) = 0.8241
Thus, the approximate value of the first intersection point is x = 0.8241.
Now we need to find the second intersection point. By graphing the two functions, we can see that there is another intersection point in the interval [2,3]. Let's choose x = 2.5 as our starting value.
x_0 = 2.5
x_1 = x_0 - h(x_0)/h'(x_0) = 2.3214
x_2 = x_1 - h(x_1)/h'(x_1) = 2.3111
x_3 = x_2 - h(x_2)/h'(x_2) = 2.3111
Thus, the approximate value of the second intersection point is x = 2.3111.
Therefore, the points of intersection of the two graphs are approximately (0.8241, f(0.8241)) and (2.3111, f(2.3111)), where f(x) = x^6.
Hope this helps you in some way! I'm sorry if it doesn't. If you need more help, ask me! :]