Respuesta :

f'(x) = 5sin(x) + 5x*cos(x) = 0. 

Then, setting f'(x) = 0 gives: 
5sin(x) + 5x*cos(x) = 0 ==> tan(x) + x = 0. 

Then, use Newton's Method to solve tan(x) + x = 0 on 0 <= x <= π. This solves to get x ≈ 2.028758. Using the Second Derivative Test, you can show that this gives a maximum. Therefore, the required maximum value is: 
f(2.028758) = 2(2.028758)sin(2.028758) ≈ 3.639411........google told me this

Answer:

x = 2.028766

Explanation:

Maximum is found when f'(x) = 0

Given:

f(x) = 3x sin (x)

f'(x) = 3x cos(x) + 3sin(x)

Newton's equation to find roots is:

[tex] x_{i+1} = x_{i} - \frac{g(x_{i})}{g'(x_{i})} [/tex]

where:

[tex] x_{i} [\tex] is the initial guess  

[tex] x_{i+1} [\tex] is the next guess

[tex] g(x_{i}) [\tex] is the original function  evaluated at [tex] x_{i} [\tex]

[tex] g'(x_{i}) [\tex] is the first derivative of the original function  evaluated at [tex] x_{i} [\tex]

Let's call g(x) = f'(x), then:

g'(x) = −3x sin(x) + 6cos(x)

Choosing x = 1.6 as initial guess:

[tex] x_{1} = 1.6 - \frac{3(1.6) cos(1.6) + 3sin(1.6)}{-3(1.6) sin(1.6) + 6cos(1.6)} [/tex]

[tex] x_{1} = 2.174799[\tex]

[tex] x_{2} = 2.174779 - \frac{3(2.174779) cos(2.174779) + 3sin(2.174779)}{-3(2.174779) sin(2.174779) + 6cos(2.174779)} [/tex]

[tex] x_{2} = 2.033956[\tex]

[tex] x_{3} = 2.033956 - \frac{3(2.033956) cos(2.033956) + 3sin(2.033956)}{-3(2.033956) sin(2.033956) + 6cos(2.033956)} [/tex]

[tex] x_{3} = 2.028766[\tex]

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