Respuesta :

Determine the x-value at which the slope of the tangent line to the curve of f(x) = 6x*sin x is zero (horiz. tan. line).

f '(x) = 6[x*cos x+sin x * 1] = 0.  This, x*cos x + sin x = 0.  

Create the new function g(x) = x*cos x + sin x.  We want to find its roots/zeros.

Use the following formula (Newton's Method) to obtain roots/zeros:

                         f(n) 
x(n+1) = x(n) - -------                                                                                                                       f '(n)

Then for the given function's derivative, x*cos x + sin x,  we get

                         x*cos x + sin x                            x*cos x + sin x
x(n+1) = x(n) - -------------------------------- = x(n) - ------------------------
                         -x*sinx + cos x + cos x                -x*sinx + 2*cos x

Let the first (guessed) root be pi/6; this is between 0 and pi, as required.

                                                       (pi/6)*cos(pi/6) + sin(pi/6)
then the next root would be  pi/6 - ----------------------------------------
                                                        -(pi/y)*sin(pi/6) + 2*cos(pi/6)
                                 (pi/6)[sqrt(3)/2] + 1/2
which equals pi/6 - ----------------------------------------
                                  -(pi/6)*(1/2) + 2sqrt(3)/2

Lot of calculations!  Having set up this application of Newton's Method, I'm hoping you can continute this process and find one or more roots to 6 places.

Questions?  Please ask.

The absolute value of the function [tex]f\left( x \right) = 6x \times \left( {\sin x} \right)[/tex] is [tex]\boxed{10.918234}.[/tex]

Further explanation:

Given:

The function is [tex]f\left( x \right) = 6x \times \left( {\sin x} \right).[/tex]

The interval for [tex]x[/tex] is [tex]0 \leqslant x \leqslant \pi.[/tex]

Explanation:

The given function can be expressed as follows,

[tex]f\left( x \right) = 6x \times \left( {\sin x} \right)[/tex]

Differentiate the above equation with respect to [tex]x.[/tex]

[tex]\begin{aligned}\frac{d}{{dx}}f\left( x \right)&= \frac{d}{{dx}}\left( {6x \times \sin x} \right)\\&= 6x \times\frac{d}{{dx}}\left( {\sin x} \right) + \sin x \times \frac{d}{{dx}}\left( {6x} \right)\\&= 6x\left( {\cos x} \right) + 6\sin x \\&= 6\left[ {x\cos x + \sin x} \right]\\\end{aligned}[/tex]

Substitute [tex]\dfrac{d}{{dx}}f\left( x \right) = 0.[/tex]

[tex]\begin{aligned}\frac{d}{{dx}}f\left( x \right) &= 0\\6\left[ {x\cos x + \sin x} \right] &= 0\\x\cos x + \sin x &= 0\\x\cos x&= - \sin x\\x&= \frac{{ - \sin x}}{{\cos x}}\\ x&=-\tanx\\\end{aligned}[/tex]

[tex]x[/tex] can take the value between o to [tex]\pi.[/tex]

The value of the function at [tex]0[/tex] can be obtained as follows,

[tex]x = 0, x = 2.029[/tex] and [tex]x = \pi.[/tex]

The value of the function at [tex]x=0.[/tex]

[tex]\begin{aligned}f\left(0\right)&= 6 \times 0 \times \sin 0\\&= 0\\\end{aligned}[/tex]

The value of the function at [tex]x=\pi.[/tex]

[tex]\begin{aligned}f\left(\pi\right)&= 6 \times \pi  \times \sin \pi\\&= 6\pi\times0\\ &= 0\\\end{aligned}[/tex]

The value of the function at [tex]x=2.029.[/tex]

[tex]\begin{aligned}f\left({2.029} \right)&= 6 \times 2.029 \times \sin \left( {2.029} \right) \\&= 10.918234\\\end{aligned}[/tex]

The absolute value of the function [tex]f\left( x \right) = 6x \times \left( {\sin x} \right)[/tex] is [tex]\boxed{10.918234}.[/tex]

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function https://brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Derivatives

Keywords: Derivative, value of x, function, differentiate, minimum value, dy, Newton’s method, six decimal places, function, f(x)=6xsinx, 0 to pi, absolute maximum, value of thre function.

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