Newton's method is an iterative technique used to approximate the roots of a function. The general formula is:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Given function:

$f(x) = (x+4)^3 - e^{1.92x} + \frac{5\cos x}{3} - 9 - 9$

$f(x) = (x+4)^3 - e^{1.92x} + \frac{5\cos x}{3} - 18$

Step 1: Compute the derivative $f'(x)$

We differentiate each term separately:

1. $\frac{d}{dx} [(x+4)^3] = 3(x+4)^2$
2. $\frac{d}{dx} [e^{1.92x}] = 1.92 e^{1.92x}$
3. $\frac{d}{dx} \left(\frac{5\cos x}{3}\right) = \frac{5}{3}(-\sin x) = -\frac{5}{3} \sin x$
4. $\frac{d}{dx} (-18) = 0$

So,

$f'(x) = 3(x+4)^2 - 1.92 e^{1.92x} - \frac{5}{3} \sin x$

Step 2: Evaluate $f(x)$ and $f'(x)$ at $x_0 = 3.2$

We substitute $x_0 = 3.2$ into $f(x)$ and $f'(x)$.

Compute $f(3.2)$:

$(3.2+4)^3 = (7.2)^3 = 373.248$

$e^{1.92(3.2)} = e^{6.144} \approx 469.7$

$\frac{5\cos(3.2)}{3} = \frac{5 \times (-0.998)}{3} \approx \frac{-4.99}{3} \approx -1.663$

$f(3.2) = 373.248 - 469.7 - 1.663 - 18$

$f(3.2) \approx 373.248 - 469.7 - 1.663 - 18 = -116.115$

Compute $f'(3.2)$:

$3(7.2)^2 = 3(51.84) = 155.52$

$-1.92 e^{6.144} = -1.92 \times 469.7 \approx -902.7$

$-\frac{5}{3} \sin(3.2) = -\frac{5}{3} \times (-0.06) = \frac{5 \times 0.06}{3} \approx 0.1$

$f'(3.2) = 155.52 - 902.7 + 0.1$

$f'(3.2) \approx 155.52 - 902.7 + 0.1 = -747.08$

Step 3: Compute the next approximation

$x_1 = 3.2 - \frac{-116.115}{-747.08}$

$x_1 = 3.2 + 0.1554$

$x_1 \approx 3.3554$

Step 4: Repeat if needed

If higher accuracy is required, we continue iterating using $x_1 = 3.3554$ in the same process.

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Would you like me to continue to more iterations or clarify anything?

Related Questions:

1. What are the conditions for Newton’s Method to converge?
2. What happens when Newton’s Method encounters a horizontal tangent?
3. How can you improve the accuracy of Newton’s Method?
4. What are alternative methods to solve non-linear equations?
5. Can Newton’s Method be used to find complex roots?

Tip: Always check the derivative to ensure it's not too small, as division by very small numbers can cause numerical instability.