Use Newton's Method to approximate the zero(s) of the function. Continue the itarations until two suocessive approximations differ by less than 0.001 . Then find the zero(s) using a graphing utility and compare the results. (Round your answers to four decimal places.) $$ f(x)=x^{3}-\cos (x) $$ Newton's method: $$ \quad x=\square $$ Graphing utlity: $$ \quad x= $$

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$$
\textbf{Step 1. Write the function and its derivative:}
$$
$$
f(x)=x^3-\cos x,\quad f'(x)=3x^2+\sin x.
$$

$$
\textbf{Step 2. Newton’s Method formula:}
$$
$$
x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
$$

$$
\textbf{Step 3. Choose an initial guess.}
$$
We note that $$ f(0)= -\cos (0)=-1 $$ and
$$
f(1)=1^3-\cos 1\approx1-0.5403=0.4597,
$$
so a sign change occurs between 0 and 1. We choose:
$$
x_0=0.5.
$$

$$
\textbf{Iteration 1 ($$x_0=0.5$$):}
$$
$$
f(0.5)=0.5^3-\cos 0.5=0.125-0.8776\approx-0.7526.
$$
$$
f'(0.5)=3(0.5)^2+\sin 0.5=0.75+0.4794\approx1.2294.
$$
$$
x_1=0.5-\frac{-0.7526}{1.2294}\approx0.5+0.6119=1.1119.
$$

$$
\textbf{Iteration 2 ($$x_1\approx1.1119$$):}
$$
First, compute:
$$
f(1.1119)=1.1119^3-\cos 1.1119\approx1.3739-0.4430\approx0.9309,
$$
$$
f'(1.1119)=3(1.1119)^2+\sin 1.1119\approx3(1.2362)+0.8966\approx3.7086+0.8966=4.6052.
$$
$$
x_2=1.1119-\frac{0.9309}{4.6052}\approx1.1119-0.2021=0.9098.
$$

$$
\textbf{Iteration 3 ($$x_2\approx0.9098$$):}
$$
$$
f(0.9098)=0.9098^3-\cos 0.9098\approx0.7526-0.6126\approx0.1400,
$$
$$
f'(0.9098)=3(0.9098)^2+\sin 0.9098\approx3(0.8278)+0.7890\approx2.4834+0.7890=3.2724.
$$
$$
x_3=0.9098-\frac{0.1400}{3.2724}\approx0.9098-0.0428=0.8670.
$$

$$
\textbf{Iteration 4 ($$x_3\approx0.8670$$):}
$$
$$
f(0.8670)=0.8670^3-\cos 0.8670\approx0.6521-0.6466\approx0.0055,
$$
$$
f'(0.8670)=3(0.8670)^2+\sin 0.8670\approx3(0.7517)+0.7610\approx2.2551+0.7610=3.0161.
$$
$$
x_4=0.8670-\frac{0.0055}{3.0161}\approx0.8670-0.0018=0.8652.
$$

The difference $$|x_4-x_3|\approx0.0018$$ is still slightly above the $$0.001$$ threshold.

$$
\textbf{Iteration 5 ($$x_4\approx0.8652$$):}
$$
$$
f(0.8652)=0.8652^3-\cos 0.8652\approx0.6486-0.6479\approx0.0007,
$$
$$
f'(0.8652)=3(0.8652)^2+\sin 0.8652\approx2.2456+0.7610\approx3.0066.
$$
$$
x_5=0.8652-\frac{0.0007}{3.0066}\approx0.8652-0.0002=0.8650.
$$

Since $$|x_5-x_4|\approx0.0002<0.001$$, we stop the Newton iterations.

$$
\boxed{ \text{Newton's Method: } x\approx0.8650. }
$$

$$
\textbf{Step 4. Graphing Utility Comparison:}
$$

Using a graphing utility to solve $$ x^3-\cos x=0 $$ also gives 
$$
\boxed{ x\approx0.8650 }.
$$
Newton's Method approximation: $$ x \approx 0.8650 $$ Graphing Utility result: $$ x \approx 0.8650 $$

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