The graph of the function $f( x) =\sin x+x^{2}-1$ is shown in Figure 14.

Figure 14 $f(x) = \sin x + x^{2}-1$

Solution (a) The function $f( x) =\sin x+x^{2}-1$ is continuous on its domain, all real numbers, so it is continuous on the closed interval $[0,1] .$ Since $f( 0) =-1$ and $f(1) =\sin 1\approx 0.841$ have opposite signs, the Intermediate Value Theorem guarantees that $f$ has a zero in the interval $(0,1) .$

(b) We begin by finding $f^\prime ( x) =\cos x+2x.$ To use Newton’s Method with a graphing utility, we enter $$x-\dfrac{\sin x+x^{2}-1}{\cos x+2x} \qquad {\color{#0066A7}{x-\dfrac{{ f}({x}) }{{ f^\prime }({x})}}}$$

Figure 15

into the $Y= \hbox{editor}$, as shown in Figure 15. We create a table by entering the initial value $0.5$ in the $X$ column. The graphing utility computes $$0.5-\dfrac{\sin 0.5+0.5^{2}-1}{\cos 0.5+2(0.5) }=0.64410789$$

and displays $0.64411$ in column $Y_{1}$ next to $0.5.$ The value $Y_{1}$ is the second approximation $c_{2}$ that we use in the next iteration. That is, we enter $0.64410789$ in the $X$ column of the next row, and the new entry in column $Y_{1}$ is the third approximation $c_{3}.$ We repeat the process until we obtain the desired approximation. The fourth approximation to the zero of $f$ is $0.63673$, as shown in Figure 16.