Please help me to solve this problem!

Question

Use Newton's method to find the absolute maximum value of the function

f(x) = 5x cos(x),    0 ≤ x ≤ 𝜋,

correct to six decimal places.

please help me to solve this problem. I've got the wrong answer.

Answer 1

f(x) = 5xcos(x) {0≤x≤π}

To find the absolute maximum of the function on that interval, check the function value at any critical numbers and at the endpoints of the closed interval.

f(0) = 0

f(π) = 5π(-1) = -5π

Critical numbers are found be setting 1st derivative equal to 0.

f'(x) = 5x(-sinx) + 5cos(x) = 5(cosx - xsinx)

When does the derivative equal zero?

cosx - xsinx = 0

Here is where Newton's method comes into play.

q(x) = cos(x) - xsin(x)

q'(x) = -sin(x) -xcos(x) -sin(x) = -2sin(x) -xcos(x)

Using Newton's Method to solve this equation on the closed interval reveals a root of approximately:

r₁=0.860333589019

And f(r₁)=2.8054817.

This latter is the absolute maximum value of the function on the given closed interval.

Here is a graph where the above values were determined, but with different notation for the functions in question.

desmos.com/calculator/bxfveisovw

Answer 2

Like in the other NM question here we can define f(x) = 5x cos x. g(x) = f'(x) and h(x) = x - g(x)/g'(x).

Creating the table and replacing y in the table with h(x₁), the table performs the NM step for us.

Using the graph of f(x) for initial guess of 0.86, putting this in first row of table gives x₂ = 0.86033362 in the second column. This becomes x for the next (second) row.

Taking this to the second row gives in the second column x₃ = 0.86033359. Repeating this in third row we get the same value for x₄ = 0.86033359, and so we stop because the precision offered is enough for us to get to six decimals x = 0.860334 (Realized that they don't want x to six decimals, see below.)

Then asking for f(0.86033359) gives 2.80548169096 or 2.805482 to six decimals.

Hope this helps!