- Write the Newton update using the starting value, the function, and its derivative. - Evaluate the function and derivative at the same starting input. - Think about where the tangent line through the current graph point meets the x-axis.

1. Differentiate: \(f'(x)=2x\). 2. Newton's update is \(x_1=x_0-\frac{f(x_0)}{f'(x_0)}\). 3. Since \(f(1.5)=0.25\) and \(f'(1.5)=3\), \(x_1=1.5-\frac{0.25}{3}=\frac{17}{12}\approx1.4167\). 4. Geometrically, \(x_1\) is the x-intercept of the tangent line to \(y=f(x)\) at \(x=x_0\).

\(x_1=\frac{17}{12}\approx1.4167\). It is the x-intercept of the tangent line at \(x_0=1.5\).

- A root is located where a graph meets the x-axis. - Which plotted line does Newton's method use to predict the next root estimate? - Read where that line crosses the horizontal axis.

1. The tangent at \(x_0=2\) has equation \(y=4x-6\). 2. Setting \(y=0\) gives the intercept \(x=\frac{3}{2}\). 3. Newton's method uses this intercept as the next root estimate, so \(x_1=1.5\).

\(x_1=1.5\), the x-intercept of the tangent.

- Match the starting value, function value, and slope with the three parts of the formula. - The fraction uses the function value over the derivative value. - Subtract the quotient from the starting value.

1. Substitute the supplied values into Newton's formula. 2. \(x_1=3-\frac{5}{2}=\frac{1}{2}\).

\(x_1=\frac{1}{2}\).

- Find a formula for the tangent slope first. - Evaluate both the function and its derivative at the starting value. - Watch the sign of the function value in the subtraction.

1. Differentiate: \(f'(x)=2x\). 2. At \(x_0=2\), \(f(2)=-1\) and \(f'(2)=4\). 3. Newton's update gives \(x_1=2-\frac{-1}{4}=\frac{9}{4}\).

\(x_1=\frac{9}{4}\).

- Look at the denominator of Newton's formula. - Evaluate the derivative at the starting value. - What happens when the tangent is horizontal?

1. Differentiate: \(f'(x)=3x^2\), so \(f'(0)=0\). 2. The Newton update divides by \(f'(x_0)\), which would mean dividing by zero. 3. Thus the first step is undefined for this starting value.

Because \(f'(0)=0\), Newton's update would divide by zero.

- A residual is the function value at the proposed root. - Check whether the function value is zero. - A small nonzero value can indicate a nearby root but not an exact one.

1. Substitute: \(f(1.4)=1.4^2-2=1.96-2=-0.04\). 2. An exact root would make the function value zero, so \(1.4\) is only an approximation.

The residual is \(-0.04\). No, \(1.4\) is not an exact root.

- Recall the general iteration rule for Newton's method. - Determine which derivative is needed in the formula. - Evaluate the function and its derivative at the current approximation before substituting. - Keep full precision until the final rounding step. - Geometrically, the next approximation is the x-intercept of the tangent at the current approximation.

1. Differentiate the function: \(f'(x) = 3x^2 + 2\). 2. For the first iteration, \(f(1) = 1^3 + 2 \cdot 1 - 5 = -2\) and \(f'(1) = 3 \cdot 1^2 + 2 = 5\). Therefore, \(x_1 = 1 - \frac{-2}{5} = 1.4\). 3. For the second iteration, \(f(1.4) = 1.4^3 + 2 \cdot 1.4 - 5 = 0.544\) and \(f'(1.4) = 3 \cdot 1.4^2 + 2 = 7.88\). 4. Thus, \(x_2 = 1.4 - \frac{0.544}{7.88} \approx 1.330964\). 5. Rounded to four decimal places, \(x_1 = 1.4000\) and \(x_2 \approx 1.3310\).

\(x_1 = 1.4000\) \(x_2 \approx 1.3310\)

- Use the Newton iteration \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). - Keep all values as fractions so the result remains exact. - When dividing by a fraction, multiply by its reciprocal. - Check whether each final fraction can be reduced. - Geometrically, the next approximation is the x-intercept of the tangent at the current approximation.

1. Let \(f(x) = x^3 - 3x^2 + 1\). Then \(f'(x) = 3x^2 - 6x\). 2. At \(x_0 = 3\), \(f(3) = 1\) and \(f'(3) = 9\). Therefore, \(x_1 = 3 - \frac{1}{9} = \frac{26}{9}\). 3. Evaluate the function at \(x_1\): \(f\left(\frac{26}{9}\right) = \frac{53}{729}\). 4. Evaluate the derivative at \(x_1\): \(f'\left(\frac{26}{9}\right) = \frac{208}{27}\). 5. Apply the Newton update again: \(x_2 = \frac{26}{9} - \frac{\frac{53}{729}}{\frac{208}{27}} = \frac{26}{9} - \frac{53}{5616} = \frac{16171}{5616}\). 6. The numerator and denominator have no common factor, so the fraction is in simplest form.

\(x_1 = \frac{26}{9}\) \(x_2 = \frac{16171}{5616}\)

- Write the general iteration formula for Newton's method. - Find the derivative needed in the formula. - Use the value of \(x_1\) as the input for the second iteration. - Keep intermediate values as fractions or retain enough decimal precision. - Geometrically, the next approximation is the x-intercept of the tangent at the current approximation.

1. Differentiate: \(f'(x) = 3x^2 + 2x + 1\). 2. At \(x_0 = 1\), \(f(1) = -2\) and \(f'(1) = 6\). Therefore, \(x_1 = 1 - \frac{-2}{6} = \frac{4}{3} \approx 1.3333\). 3. At \(x_1 = \frac{4}{3}\), \(f\left(\frac{4}{3}\right) = \frac{13}{27}\) and \(f'\left(\frac{4}{3}\right) = 9\). 4. The second approximation is \(x_2 = \frac{4}{3} - \frac{\frac{13}{27}}{9} = \frac{311}{243} \approx 1.2798\).

\(x_1 \approx 1.3333\); \(x_2 \approx 1.2798\)

- Each new Newton value is where the current tangent meets the x-axis. - Use the derivative to write the numerical update for the first tangent point. - Carry the unrounded first approximation into the second update.

1. Newton's method uses the x-intercept of the tangent at the current point as the next approximation. The graph gives estimates \(x_1\approx1.7\) and \(x_2\approx1.67\). 2. Differentiate: \(f'(x)=1.5x^2+1\). At \(x_0=2\), \(f(2)=2\) and \(f'(2)=7\), so \(x_1=2-\frac27=\frac{12}{7}\approx1.7143\). 3. At \(x_1=\frac{12}{7}\), \(f(x_1)=\frac{80}{343}\) and \(f'(x_1)=\frac{265}{49}\). 4. Therefore \(x_2=\frac{12}{7}-\frac{\frac{80}{343}}{\frac{265}{49}}=\frac{620}{371}\approx1.6712\).

a) From the tangent x-intercepts, \(x_1\approx1.7\) and \(x_2\approx1.67\). b) \(x_1=\frac{12}{7}\approx1.7143\); \(x_2=\frac{620}{371}\approx1.6712\).

- Begin by writing the Newton iteration formula. - Differentiate the function before substituting values. - Carefully place \(f(x_n)\) and \(f'(x_n)\) in the correct parts of the fraction. - Use a calculator for the powers, but record the main intermediate values. - Geometrically, the next approximation is the x-intercept of the tangent at the current approximation.

1. Differentiate: \(f'(x) = 4x^3 + 2\). 2. At \(x_0 = 1\), \(f(1) = -2\) and \(f'(1) = 6\). Hence, \(x_1 = 1 - \frac{-2}{6} = \frac{4}{3} \approx 1.3333\). 3. At \(x_1 = \frac{4}{3}\), \(f\left(\frac{4}{3}\right) = \frac{67}{81}\) and \(f'\left(\frac{4}{3}\right) = \frac{310}{27}\). 4. The second approximation is \(x_2 = \frac{4}{3} - \frac{\frac{67}{81}}{\frac{310}{27}} = \frac{391}{310} \approx 1.2613\).

\(x_1 \approx 1.3333\) \(x_2 \approx 1.2613\)

- Think about how Newton's method uses the tangent line to improve an estimate of a zero. - Check the derivative before each substitution. - Use each new approximation as the input for the next iteration. - Follow the requested rounding only after retaining enough precision in the calculation. - Geometrically, the next approximation is the x-intercept of the tangent at the current approximation.

1. Define \(f(x) = x^3 - 4x + 1\). Its derivative is \(f'(x) = 3x^2 - 4\). 2. At \(x_0 = 2\), \(f(2) = 1.0000\) and \(f'(2) = 8.0000\). Therefore, \(x_1 = 2 - \frac{1}{8} = 1.8750\). 3. At \(x_1 = 1.8750\), \(f(1.8750) = 0.091796875\) and \(f'(1.8750) = 6.546875\). 4. Thus, \(x_2 = 1.8750 - \frac{0.091796875}{6.546875} \approx 1.8610\).

\(x_1 = 1.8750\) \(x_2 \approx 1.8610\)

- Find the first derivative of the polynomial. - Write Newton's update for this function. - Use each new approximation as the input for the next iteration. - Keep more than three decimal places during the calculations. - Use the tangent's x-intercept as a visual check on the first update.

1. Differentiate: \(f'(x)=3x^2+4x\). 2. The Newton iteration is \(x_{n+1}=x_n-\frac{x_n^3+2x_n^2-7}{3x_n^2+4x_n}\). 3. Starting with \(x_0=1.5\), the successive approximations are \(x_1\approx1.431372549\), \(x_2\approx1.428821161\), and \(x_3\approx1.428817702\). 4. The approximations \(x_2\) and \(x_3\) both round to \(1.429\), so the first three decimal places have stabilized.

The real zero is \(x\approx1.429\).

- Find the first derivative and substitute it into Newton's formula. - Begin with the given value \(x_0=1.5\). - Continue until consecutive approximations agree to three decimal places.

1. Differentiate: \(f'(x)=4x^3-5\). 2. Newton's iteration is \(x_{n+1}=x_n-\frac{x_n^4-5x_n+1}{4x_n^3-5}\). 3. Starting with \(x_0=1.5\), the iterations are \(x_1\approx1.669117647\), \(x_2\approx1.638533856\), \(x_3\approx1.637307703\), and \(x_4\approx1.637305775\). 4. The values stabilize to three decimal places at \(1.637\).

The zero in \([1, 2]\) is \(x\approx1.637\).

- The first graphical approximation is the x-intercept of the shown tangent. - For later steps, use each approximation as the input to the next Newton update. - Stop when consecutive values round to the same hundredth.

1. The tangent at \(x_0=0.5\) crosses the x-axis near \(x=0.67\), so the graph suggests \(x_1\approx0.67\). 2. Differentiate: \(f'(x)=3x^2-6x\). Newton's update gives \(x_1=\frac23\approx0.666667\). 3. Continuing gives \(x_2\approx0.652778\) and \(x_3\approx0.652704\). 4. The last two approximations agree to two decimal places, so the solution is \(x\approx0.65\).

a) \(x_1\approx0.67\) b) The solution approached from \(x_0=0.5\) is \(x\approx0.65\).

- Use the given equation as the function whose zero you need. - Differentiate that function for Newton's update. - Keep several decimal places during the first and second iterations.

1. Let \(h(x)=3x^5+x-5\). Then \(h'(x)=15x^4+1\). 2. At \(x_0=1\), \(h(1)=-1\) and \(h'(1)=16\). Therefore, \(x_1=1-\frac{-1}{16}=1.0625\). 3. At \(x_1=1.0625\), \(h(1.0625)\approx0.124743\) and \(h'(1.0625)\approx20.116440\). 4. Thus, \(x_2=1.0625-\frac{0.124743}{20.116440}\approx1.056299\). 5. To three decimal places, the closest point occurs at \(x\approx1.056\).

After two iterations, \(x_2\approx1.056\).

- The next Newton approximation is where the current tangent meets the x-axis. - Read the x-intercept of tangent line \(t\) from the graph. - Substitute \(x_1\) into both the function and its first derivative. - Use those two values in Newton's formula.

1. The tangent at \(x_0=3\) meets the x-axis at \(x=2\). Therefore, \(x_1=2\). Graphically, each Newton approximation is the x-intercept of the tangent at the previous approximation. 2. Evaluate the function: \(f(2)=\frac{1}{6}\cdot8-\frac{2}{3}\cdot4+\frac{3}{2}\cdot2-1=\frac{2}{3}\). 3. Differentiate: \(f'(x)=\frac{1}{2}x^2-\frac{4}{3}x+\frac{3}{2}\). Thus, \(f'(2)=2-\frac{8}{3}+\frac{3}{2}=\frac{5}{6}\). 4. Apply Newton's formula: \(x_2=2-\frac{\frac{2}{3}}{\frac{5}{6}}=2-\frac{4}{5}=1.2\).

a) \(x_1=2\); it is the x-intercept of the tangent at \(x_0=3\). b) \(f(2)=\frac{2}{3}\) and \(f'(2)=\frac{5}{6}\) c) \(x_2=1.2\)

- The starting value is the x-coordinate of the point where the tangent touches the curve. - The tangent’s x-intercept is the next approximation. - Use \(x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime}(x_n)}\).

1. The tangent touches the graph at \(x_0=4\), so the starting value is \(4\). 2. The tangent crosses the x-axis at about \(x_1\approx3.1\). 3. Since \(f^{\prime}(x)=x\), \(f(4)=0.5\cdot4^2-4.5=3.5\) and \(f^{\prime}(4)=4\). 4. Newton’s formula gives \(x_1=x_0-\frac{f(x_0)}{f^{\prime}(x_0)}=4-\frac{3.5}{4}=3.125\), which agrees with the graphical estimate.

a) \(x_0=4\) and \(x_1\approx3.1\) b) \(x_1=4-\frac{3.5}{4}=3.125\)

- Follow the shown tangent line to its x-intercept. - For the calculation, use the function value and derivative at \(x=2.5\).

1. At \(x_0=2.5\), \(f(2.5)=6.625\). From the graph, the tangent crosses the x-axis slightly to the right of \(x=2\), at about \(x=2.1\). 2. Differentiate: \(f'(x)=3x^2-4\), so \(f'(2.5)=14.75\). 3. Apply Newton's formula: \(x_1=2.5-\frac{6.625}{14.75}=\frac{5}{2}-\frac{53}{118}=\frac{121}{59}\approx2.0508\). 4. This agrees with the graphical estimate.

a) The tangent crosses the x-axis at approximately \(2.1\). b) \(x_1=\frac{121}{59}\approx2.051\)

- Follow the shown tangent to where it meets the x-axis. - For the exact calculation, evaluate the function and its derivative at the stated starting value. - Compare the algebraic update with the geometry of the tangent line.

1. From the figure, the tangent crosses the x-axis near \(x=-2.5\), so \(x_1\) should be about \(-2.5\). 2. Differentiate: \(f'(x)=0.75x^2-1\). At \(x_0=-3\), \(f(-3)=-2.75=-\frac{11}{4}\) and \(f'(-3)=5.75=\frac{23}{4}\). 3. Newton's formula gives \(x_1=-3-\frac{-11/4}{23/4}=-3+\frac{11}{23}=-\frac{58}{23}\approx-2.522\). 4. The Newton update is obtained by taking the tangent at \((x_0,f(x_0))\) and using the x-coordinate of that tangent's x-intercept.

a) From the graph, \(x_1\approx-2.5\). Exactly, \(x_1=-\frac{58}{23}\approx-2.522\). b) Newton's next approximation is the x-intercept of the tangent at the current approximation.

- Newton's update needs a nonzero tangent slope. - Check the slope at each proposed starting value. - For the valid start, use the function value and slope to compute the intercept.

1. Differentiate: \(f'(x)=3x^2-3\). 2. At \(x=1\), \(f'(1)=0\), so Newton's formula is undefined. 3. At \(x=0\), \(f(0)=1\) and \(f'(0)=-3\). 4. The valid update is \(x_1=0-\frac{1}{-3}=\frac{1}{3}\).

Start at \(x_0=0\); the next estimate is \(x_1=\frac{1}{3}\). Starting at \(1\) would divide by zero.

- Round each listed estimate to the requested number of decimal places. - Compare neighboring estimates, not every estimate with the starting value. - A stopping rule is about the listed estimates; it is not a proof of exactness.

1. Rounding the estimates to three decimals gives \(1.500\), \(1.417\), \(1.414\), and \(1.414\). 2. The first consecutive matching pair occurs at steps \(2\) and \(3\). 3. The method stops after step \(3\) under the stated rule and reports \(1.414\).

Stop after step \(3\) and report \(1.414\).

- Make each update from its own starting value. - A residual measures how far the resulting function value is from zero. - Compare the absolute residuals; do not assume the larger starting value is better.

1. With \(x_0=1\), \(f(1)=-1\) and \(f'(1)=2\), so \(x_1=1-\frac{-1}{2}=\frac{3}{2}\). 2. Its absolute residual is \(|f(\frac{3}{2})|=|\frac{9}{4}-2|=\frac{1}{4}\). 3. With \(x_0=3\), \(f(3)=7\) and \(f'(3)=6\), so \(x_1=3-\frac{7}{6}=\frac{11}{6}\). 4. Its absolute residual is \(|f(\frac{11}{6})|=|\frac{121}{36}-2|=\frac{49}{36}\). 5. Since \(\frac{1}{4}<\frac{49}{36}\), the start at \(1\) performs better on this one-step residual comparison.

Starting at \(x_0=1\) gives the smaller absolute residual: \(\frac{1}{4}\) versus \(\frac{49}{36}\) for \(x_0=3\).

- Use the function value and derivative separately at each new estimate. - Do not assume the estimates must move in one direction. - What happens if an update returns you to a starting value already used?

1. Differentiate: \(f'(x)=3x^2-2\). 2. At \(x_0=0\), \(f(0)=2\) and \(f'(0)=-2\), so \(x_1=0-\frac{2}{-2}=1\). 3. At \(x_1=1\), \(f(1)=1\) and \(f'(1)=1\), so \(x_2=1-\frac{1}{1}=0\). 4. The next step repeats the first, so the iterates cycle between \(0\) and \(1\), neither of which is a root.

\(x_1=1\) and \(x_2=0\). The iteration cycles \(0,1,0,1,\dots\), so it does not approach a root from this starting value.

- What could prevent a Newton iteration from being calculated at a starting value? - Which local feature of the function controls where the next approximation lands? - How can you verify that your calculated next input follows the method?

1. The derivative is \(f'(x)=3x^2\), so \(f'(0)=0\). Newton's formula would divide by zero at \(x_0=0\). 2. For \(x_0=1\), \(f(1)=-1\) and \(f'(1)=3\). 3. The iteration gives \(x_1=1-\frac{-1}{3}=\frac{4}{3}\).

a) \(f'(0)=0\), so the Newton step is undefined. b) \(x_1=\frac{4}{3}\).

- Take one Newton step from the stated starting point. - Distinguish the size of the function value from the distance between the input and the actual root. - This quadratic has a positive root that you can check exactly.

1. Since \(f'(x)=2x\), Newton's formula gives \(x_1=4-\frac{16-9}{8}=\frac{25}{8}=3.125\). 2. The function value is \(f(x_1)=\frac{625}{64}-9=\frac{49}{64}\), whose absolute value is less than \(1\). 3. The positive root is \(3\), and \(|x_1-3|=\frac{1}{8}=0.125>0.1\). The claim is false: the stated residual test does not establish the claimed input accuracy.

The claim is false. \(x_1=3.125\) and \(|f(x_1)|=\frac{49}{64}<1\), but \(|x_1-3|=0.125>0.1\).

- Read the formula in C1 and replace the cell reference B1 with \(x\). - Compare the spreadsheet formulas with the Newton iteration \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). - The initial value is the first number used in the calculation. - To determine B2, evaluate the formulas in the first row in order.

1. The formula in C1 is based on \(0.5x^2 - 4\), so \(f(x) = 0.5x^2 - 4\). Cell B1 gives the initial value \(x_0 = 4\). 2. Cell E1 calculates the correction term \(\frac{f(x_n)}{f'(x_n)}\). Cell F1 subtracts that correction from the current approximation to produce \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). 3. With \(x_0 = 4\), C1 gives \(f(4) = 0.5 \cdot 4^2 - 4 = 4\), and D1 gives \(f'(4) = 4\). Therefore, E1 is \(4 \div 4 = 1\), F1 is \(4 - 1 = 3\), and B2 displays \(3\).

1. \(f(x) = 0.5x^2 - 4\); \(x_0 = 4\) 2. E1 calculates \(\frac{f(x_n)}{f'(x_n)}\); F1 calculates the next approximation \(x_{n+1}\). 3. Cell B2 displays \(3\).

- Identify the derivative information Newton's method needs at the current approximation. - Build the function-value and derivative-value cells from the formulas for \(f\) and its derivative, using B1 for the current input. - Use the values already computed in the row to build the next-approximation cell. - Use your value of \(x_1\) as the input for the second iteration.

1. Differentiate the polynomial: \(f'(x)=3x^2-2\). 2. The spreadsheet formulas are C1: `=B1^3-2*B1-5`; D1: `=3*B1^2-2`; E1: `=B1-C1/D1`. 3. At \(x_0=2\), \(f(2)=-1\) and \(f'(2)=10\). Thus, \(x_1=2-\frac{-1}{10}=2.1\). 4. At \(x_1=2.1\), \(f(2.1)=0.061\) and \(f'(2.1)=11.23\). Therefore, \(x_2=2.1-\frac{0.061}{11.23}\approx2.09457\).

1. \(f'(x)=3x^2-2\) 2. C1: `=B1^3-2*B1-5`; D1: `=3*B1^2-2`; E1: `=B1-C1/D1` 3. \(x_1=2.1\); \(x_2\approx2.09457\)

- Use function values with opposite signs to establish existence of a zero. - Examine the first derivative to determine whether the function is monotonic. - Write the Newton iteration for this specific function. - Continue until the requested decimal places stop changing. - Geometrically, the first Newton update is where the shown tangent crosses the x-axis.

1. Since \(f\) is a polynomial, it is continuous. Also, \(f(1) = -2\) and \(f(2) = 10\), so the Intermediate Value Theorem guarantees at least one zero in \((1, 2)\). 2. The derivative is \(f'(x) = 3x^2 + 5\), which is positive for every real \(x\). Therefore, \(f\) is strictly increasing and can have at most one zero. Together, these facts show that the zero is unique. 3. Apply \(x_{n+1} = x_n - \frac{x_n^3 + 5x_n - 8}{3x_n^2 + 5}\): \(x_1 = 1.25\), \(x_2 \approx 1.2290323\), \(x_3 \approx 1.2288603\), and \(x_4 \approx 1.2288602\). 4. The last two approximations agree to four decimal places, so the zero is \(x \approx 1.2289\).

a) A sign change from \(f(1) = -2\) to \(f(2) = 10\) proves existence, and \(f'(x) = 3x^2 + 5 > 0\) for all \(x\) proves uniqueness. b) \(x \approx 1.2289\)

- Determine the signs of the function values at the interval endpoints. - Use the derivative to analyze whether the function can cross the x-axis more than once in the interval. - Select a simple starting value inside \([0, 1]\). - Keep more precision in the iterations than the final answer requires. - Use the tangent's x-intercept to interpret the first Newton approximation geometrically.

1. The function is continuous, with \(g(0) = -2\) and \(g(1) = 2\). The Intermediate Value Theorem therefore guarantees at least one zero in \([0, 1]\). 2. Since \(g'(x) = 4x^3 + 3 > 0\) for every \(x \in [0, 1]\), the function is strictly increasing on the interval. Thus, the zero is unique. 3. Choose \(x_0 = 0.5\). Using \(x_{n+1} = x_n - \frac{x_n^4 + 3x_n - 2}{4x_n^3 + 3}\) gives \(x_1 = 0.625\), \(x_2 \approx 0.6180624\), \(x_3 \approx 0.6180340\), and \(x_4 \approx 0.6180340\). 4. Therefore, the zero is \(x \approx 0.6180\) to four decimal places.

a) The sign change between \(g(0) = -2\) and \(g(1) = 2\) proves existence. Because \(g'(x) = 4x^3 + 3 > 0\) on \([0, 1]\), \(g\) is strictly increasing there, so the zero is unique. b) Using \(x_0 = 0.5\), the zero is \(x \approx 0.6180\).

- Translate the slope condition into a zero-finding problem for a new function. - The tangent's x-intercept is the first Newton approximation. - Use the derivative of the zero-finding function for the numerical update. - Keep extra decimal places until the final rounding.

1. Differentiate: \(f'(x)=x^3-3x^2+5\), so \(F(x)=x^3-3x^2+3\). 2. From the graph, the tangent at \(x_0=2.5\) crosses the x-axis near \(x=2.53\), so \(x_1\approx2.53\). 3. Since \(F'(x)=3x^2-6x\), \(F(2.5)=-0.125\) and \(F'(2.5)=3.75\). Thus \(x_1=2.5-\frac{-0.125}{3.75}=\frac{38}{15}\approx2.5333\). 4. A second iteration gives \(x_2\approx2.5321\). Therefore, \(x^*\approx2.53\).

a) \(x_1\approx2.53\). b) \(x^*\approx2.53\).

- Create a function whose zero occurs when \(g(x)=h(x)\). - Find the derivative of the difference function. - Apply Newton's update twice. - Use the first approximation as the input for the second step. - The first tangent x-intercept gives a geometric check on your first update.

1. Define the difference function \(F(x)=g(x)-h(x)=x^3+2x-6\). Its zero gives the intersection x-coordinate. 2. Differentiate: \(F'(x)=3x^2+2\). 3. At \(x_0=1.5\), \(F(1.5)=0.375\) and \(F'(1.5)=8.75\). Thus, \(x_1=1.5-\frac{0.375}{8.75}\approx1.457142857\). 4. At \(x_1\), \(F(x_1)\approx0.00818659\) and \(F'(x_1)\approx8.36979592\). Therefore, \(x_2\approx1.456164746\).

After two iterations, \(x^*\approx1.4562\).

- Critical numbers occur where the first derivative equals zero. - Apply Newton's method to the first derivative, not to the original function. - Differentiate the function twice before writing the iteration formula. - Keep exact fractions during the iterations when possible. - When Newton's method is applied to \(g'\), each tangent is drawn on the derivative graph.

1. A local minimum occurs where \(g'(x)=0\). Differentiate: \(g'(x)=2x^3-3x^2+4x-1\) and \(g''(x)=6x^2-6x+4=6(x-0.5)^2+2.5>0\). Thus, \(g'\) is strictly increasing, and its unique zero gives the minimum. 2. Apply Newton's formula to \(g'\): \(x_{n+1}=x_n-\frac{g'(x_n)}{g''(x_n)}\). 3. With \(x_0=0\), \(g'(0)=-1\) and \(g''(0)=4\), so \(x_1=0-\frac{-1}{4}=0.25\). 4. At \(x_1=0.25\), \(g'(0.25)=-\frac{5}{32}\) and \(g''(0.25)=\frac{23}{8}\). 5. Therefore, \(x_2=0.25-\frac{-\frac{5}{32}}{\frac{23}{8}}=\frac{7}{23}\approx0.3043\).

After two iterations, the local minimum occurs at approximately \(x\approx0.304\).

- The graphs intersect where their function values are equal. - Rewrite the intersection equation as one function equal to zero. - Differentiate that difference function for Newton's formula. - Continue until the requested decimal places no longer change. - The tangent's x-intercept provides a visual interpretation of the first Newton update.

1. Define the difference function \(h(x)=f(x)-g(x)=x^3-x^2+2x+5\). An intersection occurs where \(h(x)=0\). 2. Differentiate: \(h'(x)=3x^2-2x+2\). 3. Starting with \(x_0=-1\), apply \(x_{n+1}=x_n-\frac{h(x_n)}{h'(x_n)}\). This gives \(x_1\approx-1.142857\), \(x_2\approx-1.132552\), \(x_3\approx-1.132494\), and \(x_4\approx-1.132494\). 4. The approximations have stabilized to three decimal places, so the intersection x-coordinate is \(x\approx-1.132\).

The x-coordinate of the intersection is \(x\approx-1.132\).

- Find the derivative of \(x^4-a\). - Put the terms in Newton's formula over a common denominator. - Keep unrounded values during the iterations. - A fourth root of \(a\) is a zero of \(x^4-a\).

1. Differentiate \(f(x) = x^4-a\): \(f'(x) = 4x^3\). 2. Substitute into Newton's formula: \(x_{n+1} = x_n - \frac{x_n^4-a}{4x_n^3}\). 3. Combine the terms: \(x_{n+1} = \frac{4x_n^4-(x_n^4-a)}{4x_n^3} = \frac{3x_n^4+a}{4x_n^3} = \frac{1}{4}\left(3x_n+\frac{a}{x_n^3}\right)\). 4. For \(a=100\) and \(x_0=3\), \(x_1 = \frac{1}{4}\left(9+\frac{100}{27}\right) \approx 3.1759259\), so \(x_1 \approx 3.1759\). 5. Using the unrounded value of \(x_1\), \(x_2 = \frac{1}{4}\left(3x_1+\frac{100}{x_1^3}\right) \approx 3.1623654\), so \(x_2 \approx 3.1624\). 6. The unrounded second approximation gives \(x_2^4 \approx 100.0111\). Using the rounded value gives \(3.1624^4 \approx 100.0155\), which differs from \(100\) by about \(0.0155\).

a) \(x_{n+1} = \frac{1}{4}\left(3x_n + \frac{a}{x_n^3}\right)\) b) \(x_1 \approx 3.1759\); \(x_2 \approx 3.1624\) c) \(3.1624^4 \approx 100.0155\), so the error is about \(0.0155\).

- Convert the target volume to cubic decimeters before setting up the zero equation. - Differentiate the given volume-depth model to build Newton's update. - Keep the first approximation unrounded while calculating the second. - Each tangent's horizontal intercept gives a graphical check on the next depth.

1. Since \(120\,\text{L}=120\,\text{dm}^3\), solve \(f(h)=0.5h^3+9\pi h-120=0\) for \(h\ge0\). 2. Differentiate: \(f'(h)=1.5h^2+9\pi\). 3. At \(h_0=3\), \(f(3)=13.5+27\pi-120\approx-21.677\) and \(f'(3)=13.5+9\pi\approx41.774\). The first update gives \(h_1=3-\frac{f(3)}{f'(3)}\approx3.518907\). 4. Using the unrounded \(h_1\), the second update gives \(h_2=h_1-\frac{f(h_1)}{f'(h_1)}\approx3.491552\). 5. Rounded to three decimal places, \(h_1\approx3.519\,\text{dm}\) and \(h_2\approx3.492\,\text{dm}\).

First approximation: \(h_1\approx3.519\,\text{dm}\). Second approximation: \(h_2\approx3.492\,\text{dm}\).

- Differentiate \(x^3-a\). - Put the terms in the Newton update over a common denominator. - Substitute \(a=26\) and \(x_1=3\) into the formula one step at a time. - Keep \(x_2\) as an exact fraction, or retain enough decimal places, before calculating \(x_3\).

1. Differentiate \(f(x)=x^3-a\): \(f'(x)=3x^2\). 2. Substitute into Newton's method: \(x_{n+1}=x_n-\frac{x_n^3-a}{3x_n^2}\). 3. Simplify: \(x_{n+1}=\frac{3x_n^3-(x_n^3-a)}{3x_n^2}=\frac{2x_n^3+a}{3x_n^2}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)\). 4. For \(a=26\) and \(x_1=3\), \(x_2=\frac{1}{3}\left(6+\frac{26}{9}\right)=\frac{80}{27}\approx2.962963\). 5. Using \(x_2=\frac{80}{27}\), \(x_3=\frac{1}{3}\left(2x_2+\frac{26}{x_2^2}\right)\approx2.962496\).

a) \(x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)\) b) \(x_2=\frac{80}{27}\approx2.9630\); \(x_3\approx2.9625\)

- Create a difference function whose zeros correspond to intersections. - Check simple values to choose a starting point near the positive zero. - Apply Newton's update repeatedly until the hundredths place is stable. - Make sure the iteration approaches the intersection with \(x>0\). - The first tangent x-intercept gives a geometric check on your chosen starting value.

1. Define \(h(x)=f(x)-g(x)=x^4-2x-5\). An intersection occurs where \(h(x)=0\). 2. Differentiate: \(h'(x)=4x^3-2\). 3. Since \(h(1)=-6\) and \(h(2)=7\), the positive zero lies between \(1\) and \(2\). Choose \(x_0=1.5\). 4. Newton's method gives \(x_1\approx1.755435\), \(x_2\approx1.705272\), \(x_3\approx1.702713\), and \(x_4\approx1.702706\). 5. Therefore, the positive intersection x-coordinate is \(x\approx1.70\).

\(x\approx1.70\)

- Use the volume formula for a rectangular prism. - Triple the original volume to determine the target volume. - Expand the product and move all terms to one side. - Use the derivative of the resulting polynomial in Newton's formula. - The tangent's x-intercept provides a geometric check on the first Newton approximation.

1. The original volume is \(4\cdot5\cdot10=200\,\text{ft}^3\), so the target volume is \(600\,\text{ft}^3\). The enlarged volume is \(V(x)=(4+x)(5+x)(10+x)\). 2. Expanding gives \(V(x)=x^3+19x^2+110x+200\). Setting this equal to \(600\) gives \(x^3+19x^2+110x-400=0\). 3. Let \(f(x)=x^3+19x^2+110x-400\), so \(f'(x)=3x^2+38x+110\). Starting with \(x_0=2\), Newton's method gives \(x_1\approx2.484848\), \(x_2\approx2.457977\), and \(x_3\approx2.457891\). 4. Therefore, each dimension must be increased by approximately \(2.46\,\text{ft}\).

1. \(V(x)=(4+x)(5+x)(10+x)\) 2. \(V(x)=600\) simplifies to \(x^3+19x^2+110x-400=0\). 3. \(x\approx2.46\,\text{ft}\)

- Test simple values to locate an interval containing the solution. - Rewrite the equation with zero on one side. - Newton's iteration uses both the function and its first derivative. - Repeat the process until the requested decimal places stop changing.

1. Rewrite the equation as \(f(x)=x^3+3x-5=0\), with \(f'(x)=3x^2+3\). 2. Since \(f(1)=-1\) and \(f(2)=9\), a zero lies in \((1, 2)\). Choose \(x_0=1\). 3. Newton's method gives \(x_1\approx1.166666667\), \(x_2\approx1.154248366\), and \(x_3\approx1.154171498\). 4. The last two approximations both round to \(1.15\), so the required solution is \(x\approx1.15\).

\(x\approx1.15\)

- Write Newton's iteration for the given function. - Use the first derivative at the current approximation. - Keep enough precision in intermediate values. - Compare each result with the stabilized value.

1. Differentiate: \(f'(x)=3x^2+8x\). 2. From \(x_1=1\), \(x_2=1-\frac{-5}{11}=\frac{16}{11}\approx1.4545\), and \(x_3\approx1.3689\). 3. From \(x_1=2\), \(x_2=2-\frac{14}{28}=1.5\), and \(x_3=1.5-\frac{2.375}{18.75}\approx1.3733\). 4. The zero is approximately \(1.36523\). After two iterations, the result from \(x_1=1\) is closer, so that initial value converges slightly faster in this comparison.

a) \(x_2\approx1.4545\); \(x_3\approx1.3689\) b) \(x_2=1.5\); \(x_3\approx1.3733\) c) The initial value \(x_1=1\) is closer after two iterations.

- Use opposite endpoint signs and continuity for the existence statement. - Apply Newton's update one step at a time. - On the graph, follow each shown tangent to the x-axis and compare the two landing points.

1. Since \(f(-2)=-2\), \(f(-1)=3\), and \(f\) is continuous, the Intermediate Value Theorem guarantees a zero in \((-2,-1)\). 2. Differentiate: \(f'(x)=3x^2-2\). 3. From \(x_0=0\), \(x_1=0-\frac{2}{-2}=1\). 4. From \(x_1=1\), \(x_2=1-\frac11=0\). Repeating gives \(x_3=1\). 5. The tangent at \((0,2)\) meets the x-axis at \(x=1\), while the tangent at \((1,1)\) meets the x-axis at \(x=0\). Newton's construction therefore repeats \(0\to1\to0\to1\) and never approaches the actual zero near \(-1.7693\).

a) A zero lies in \((-2,-1)\). b) \(x_1=1\), \(x_2=0\), \(x_3=1\). c) The two tangent x-intercepts send \(0\) to \(1\) and \(1\) back to \(0\), producing a two-cycle.

- Write the Newton update and identify the function and derivative values needed. - Relate the derivative to the slope of the tangent line. - Examine the graph at \(x=1\). - Visualize the tangent at \(x=1.1\) and where it meets the x-axis.

1. Differentiate: \(f'(x)=3x^2-3\). 2. At \(x_0=2\), \(f(2)=3\) and \(f'(2)=9\). Thus, \(x_1=2-\frac{3}{9}=\frac{5}{3}\approx1.667\). 3. At \(x_1=\frac{5}{3}\), \(f\left(\frac{5}{3}\right)=\frac{17}{27}\) and \(f'\left(\frac{5}{3}\right)=\frac{16}{3}\). Therefore, \(x_2=\frac{5}{3}-\frac{\frac{17}{27}}{\frac{16}{3}}=\frac{223}{144}\approx1.549\). 4. At \(x_0=1\), \(f'(1)=0\), so Newton's formula would divide by zero. On the graph, \(x=1\) is a local minimum with a horizontal tangent, and that tangent does not intersect the x-axis. 5. Near \(x=1\), the tangent slope is small. At \(x_0=1.1\), \(f(1.1)=-0.969\) and \(f'(1.1)=0.63\), giving \(x_1\approx2.638\). Because the tangent is nearly horizontal and begins below the x-axis, its x-intercept lies far to the right.

a) \(x_1=\frac{5}{3}\approx1.667\); \(x_2=\frac{223}{144}\approx1.549\) b) \(f'(1)=0\), so Newton's formula is undefined. Graphically, the tangent at the local minimum is horizontal and does not meet the x-axis. c) Just to the right of the minimum, the tangent is almost horizontal. Its x-intercept is therefore far to the right of the nearby zero.

- Write the general Newton update. - Relate a zero derivative to the tangent line. - Find the x-intercept of the tangent at \(x_0=-1.5\). - Use the graph to see which zero lies in the path of later iterations.

1. Differentiate: \(f'(x)=0.3x^2-1.2\). 2. At \(x_0=1\), \(f(1)=-0.1\) and \(f'(1)=-0.9\). Thus, \(x_1=1-\frac{-0.1}{-0.9}=\frac{8}{9}\approx0.89\). 3. At \(x_0=2\), \(f'(2)=0\), so the Newton update is undefined. The graph has a local minimum with a horizontal tangent at that point. 4. At \(x_0=-1.5\), \(f(-1.5)=2.4625\) and \(f'(-1.5)=-0.525\). Therefore, \(x_1=-1.5-\frac{2.4625}{-0.525}=\frac{67}{21}\approx3.19\). 5. This first approximation lies to the right of the right-hand zero near \(2.93\). Subsequent tangent intercepts approach that zero from the right.

a) \(x_1=\frac{8}{9}\approx0.89\) b) \(f'(2)=0\), so the update divides by zero; the tangent is horizontal. c) \(x_1=\frac{67}{21}\approx3.19\). The iteration approaches the right-hand zero near \(2.93\).

- Apply Newton's update successively. - Track whether a previously used value reappears. - Relate each tangent line to its x-intercept. - Factor the polynomial to find the exact zeros. - Choose an initial value on the same monotonic branch as the target zero.

1. Differentiate: \(f'(x)=3x^2-5\). 2. At \(x_1=1\), \(f(1)=-4\) and \(f'(1)=-2\), so \(x_2=1-\frac{-4}{-2}=-1\). 3. At \(x_2=-1\), \(f(-1)=4\) and \(f'(-1)=-2\), so \(x_3=-1-\frac{4}{-2}=1\). The iteration is trapped in a two-cycle between \(1\) and \(-1\). 4. Graphically, the tangent at \((1, -4)\) intersects the x-axis at \(-1\), and the tangent at \((-1, 4)\) intersects it at \(1\). 5. Factoring gives \(x(x^2-5)=0\), so the positive zero is \(\sqrt{5}\approx2.236\). An initial value such as \(x_1=3\) is reasonable because it lies to the right of the positive zero and away from the nearby critical number.

a) \(x_2=-1\); \(x_3=1\). The iteration cycles and does not converge. b) The tangent at \(x=1\) has x-intercept \(-1\), and the tangent at \(x=-1\) has x-intercept \(1\), producing the cycle. c) The positive zero is \(\sqrt{5}\). A reasonable initial value is \(x_1=3\).

- Look for the points where the graph crosses the x-axis. - Use the general Newton iteration. - Consider what happens when the tangent line is horizontal. - Identify which expression in Newton's formula cannot equal zero.

1. Reading from the graph gives approximate zeros of \(-2.5\), \(0.5\), and \(1.9\). 2. Differentiate: \(f'(x)=0.6x^2-1\). 3. At \(x_0=2\), \(f(2)=0.1\) and \(f'(2)=1.4\). Therefore, \(x_1=2-\frac{0.1}{1.4}=2-\frac{1}{14}\approx1.929\). 4. An initial value where \(f'(x)=0\) makes Newton's formula undefined because it requires division by the derivative. 5. Solve \(0.6x^2-1=0\): \(x^2=\frac{5}{3}\), so \(x=\pm\sqrt{\frac{5}{3}}\approx\pm1.29\).

a) The zeros are approximately \(-2.5\), \(0.5\), and \(1.9\). b) \(x_1\approx1.929\) c) The inadmissible initial values are \(x\approx-1.29\) and \(x\approx1.29\), where \(f'(x)=0\).

- Substitute each starting value into the function and its derivative. - Compare the new value in part b) with the nearby x-intercept on the graph. - The derivative gives the slope of the tangent line. - Consider what happens in Newton's fraction when the denominator is close to zero.

1. Differentiate: \(g'(x)=1.5x^2-2\). 2. For \(x_0=2\), \(g(2)=1\) and \(g'(2)=4\). Therefore, \(x_1=2-\frac{1}{4}=1.75\). 3. For \(x_0=0\), \(g(0)=1\) and \(g'(0)=-2\). Therefore, \(x_1=0-\frac{1}{-2}=0.5\). The corresponding zero is approximately \(0.539\), so \(0.5\) is closer to it than \(0\). 4. The line is the tangent to the graph at \((x_n,g(x_n))\). Near a local extremum, the tangent slope is close to zero. An almost horizontal tangent may meet the x-axis far away, producing a large change from \(x_n\) to \(x_{n+1}\).

a) \(x_1=1.75\) b) \(x_1=0.5\), which is closer to the corresponding zero at approximately \(0.539\). c) The line is the tangent at \((x_n,g(x_n))\). Near an extremum, its slope can be close to zero, so its x-intercept can be far from \(x_n\).

- Apply Newton's formula one step at a time and retain enough decimal places. - Determine the tangent slope at \(x=0\). - Continue until consecutive approximations agree to two decimal places.

1. Differentiate: \(f'(x)=x^3-4x\). 2. Beginning with \(x_0=3\), Newton's method gives \(x_1\approx2.716666667\), \(x_2\approx2.623389823\), \(x_3\approx2.613241522\), and \(x_4\approx2.613125945\). 3. The last two approximations round to \(2.61\), so the greatest positive zero is approximately \(2.61\). 4. If \(x_0=0\), then \(f'(0)=0\), so the Newton update is undefined. Graphically, the tangent at \((0, 2)\) is horizontal and does not intersect the x-axis.

a) The greatest positive zero is \(x\approx2.61\). b) At \(x_0=0\), \(f'(0)=0\), so Newton's method would divide by zero. The tangent there is horizontal and does not meet the x-axis.

- Identify what cannot equal zero in the denominator of Newton's formula. - Locate the point reached after the first step. - Interpret a tangent slope of zero on the graph.

1. Differentiate: \(h'(x)=2x-4\). 2. The first approximation is \(x_1=3-\frac{h(3)}{h'(3)}=3-\frac{2}{2}=2\). 3. The next step requires \(h'(2)\), but \(h'(2)=0\). Therefore, the update is undefined because it would divide by zero. 4. Geometrically, \(x=2\) is the vertex of the parabola. The tangent there is horizontal and never intersects the x-axis. Also, \(h(x)=(x-2)^2+1>0\), so the function has no real zero.

a) \(x_1=2\) b) Because \(h'(2)=0\), the formula for \(x_2\) is undefined. c) The tangent at the vertex is horizontal and does not meet the x-axis; the parabola has no real zero.

- Examine the tangent when the graph is nearly flat. - Consider where an almost horizontal tangent meets the x-axis. - Evaluate both the function and its derivative at \(x_0=2.5\).

1. Differentiate: \(f'(x)=x^2-3\). The positive critical number is \(x=\sqrt{3}\approx1.732\), where the graph has a local minimum. 2. Near \(x=1.7\), the derivative is close to zero, so the tangent is nearly horizontal. Its x-intercept can be far from the starting point, causing the iteration to overshoot or move toward a different zero. 3. At \(x_0=2.5\), \(f(2.5)=\frac{125}{24}-\frac{15}{2}+1=-\frac{31}{24}\approx-1.2917\), and \(f'(2.5)=3.25\). 4. Therefore, \(x_1=2.5-\frac{-\frac{31}{24}}{3.25}\approx2.897\).

a) Near \(x=1.7\), the tangent is almost horizontal, so its x-intercept may lie far away and miss the intended zero. b) \(x_1\approx2.897\)

- Write the Newton update before substituting values. - Identify the role of the first derivative in the denominator. - On the graph, look for a local maximum or minimum. - Newton's method uses the x-intercept of the tangent line.

1. Differentiate: \(f'(x)=x^3-4x+0.5\). 2. At \(x_0=3\), \(f(3)=4.75\) and \(f'(3)=15.5\). Therefore, \(x_1=3-\frac{4.75}{15.5}\approx2.693548\), so \(x_1\approx2.694\). 3. Using the unrounded value of \(x_1\), \(f(x_1)\approx0.995861\) and \(f'(x_1)\approx9.268047\). Thus, \(x_2\approx2.586097\), so \(x_2\approx2.586\). 4. A point where \(f'(x)=0\) has a horizontal tangent, so the Newton update is undefined. For example, the visible local minimum near \(x=1.9\) is unsuitable because the tangent there is horizontal.

a) \(x_1\approx2.694\); \(x_2\approx2.586\) b) The local minimum near \(x=1.9\) is unsuitable. Its tangent is horizontal, so Newton's formula would divide by zero.

- Treat \(f'\) as the function whose zero is being found, and use \(f''\) as its derivative. - Distinguish carefully between the two graphs. - The value of \(f''\) gives the slope of the graph of \(f'\). - Check what happens when the denominator of the Newton update is zero.

1. Let \(g(x)=f'(x)=x^3-3x^2+1\). Its derivative is \(g'(x)=f''(x)=3x^2-6x\). 2. At \(x_0=3\), \(g(3)=1\) and \(g'(3)=9\). Thus, \(x_1=3-\frac{1}{9}=\frac{26}{9}\approx2.889\). 3. Using \(x_1=\frac{26}{9}\), \(g(x_1)=\frac{53}{729}\) and \(g'(x_1)=\frac{208}{27}\). Therefore, \(x_2=\frac{26}{9}-\frac{\frac{53}{729}}{\frac{208}{27}}=\frac{16171}{5616}\approx2.879\). 4. At \(x=2\), \(f''(2)=0\). Therefore, the graph of \(f'\) has a horizontal tangent there, and the Newton update for finding a zero of \(f'\) would divide by zero.

a) \(x_1\approx2.889\); \(x_2\approx2.879\) b) \(f''(2)=0\), so the graph of \(f'\) has a horizontal tangent at \(x=2\). Newton's formula is undefined there.

- Use the sign of the second derivative to determine how the first derivative changes. - A continuous strictly increasing function can cross zero at most once; combine that with its end behavior. - On the graph, the first Newton approximation is the tangent's x-intercept. - For the numerical iterations, apply Newton's method to the first derivative.

1. The first derivative is \(f'(x)=\frac{1}{2}x^3+3x+5\), and the second derivative is \(f''(x)=\frac{3}{2}x^2+3\). 2. Since \(f''(x)\ge3>0\) for every real \(x\), \(f'\) is strictly increasing. Also, \(f'(x)\to-\infty\) as \(x\to-\infty\), and \(f'(x)\to\infty\) as \(x\to\infty\). Therefore, \(f'\) has exactly one zero, so \(f\) has exactly one critical point. Because \(f''>0\), it is a minimum. 3. The tangent shown at \(x_0=-1\) crosses the x-axis near \(x=-1.33\), so the first Newton approximation is about \(-1.33\). 4. Apply Newton's method to \(f'(x)=0\): \(x_{n+1}=x_n-\frac{f'(x_n)}{f''(x_n)}\). Starting with \(x_0=-1\), the approximations are \(x_1=-\frac{4}{3}\approx-1.3333\), \(x_2\approx-1.3007\), and \(x_3\approx-1.3003\). 5. The critical point occurs at \(x\approx-1.300\).

1. Since \(f''(x)>0\), the derivative \(f'\) is strictly increasing from \(-\infty\) to \(\infty\), so it has exactly one zero. 2. The graph gives \(x_1\approx-1.33\); Newton's method gives the unique critical point at \(x\approx-1.300\), which is a minimum.

- Rewrite \(\frac{1}{x}\) as \(x^{-1}\) before differentiating. - Simplify the complex fraction by multiplying by the reciprocal of its denominator. - In part b), evaluate one iteration at a time and track the sign of \(2-bx_n\). - A sequence that moves farther from the target is diverging. - In part c), compare the tangent's x-intercept with the starting value to see why the iteration immediately moves away from the positive root.

1. Differentiate \(f(x)=x^{-1}-b\): \(f'(x)=-x^{-2}=-\frac{1}{x^2}\). 2. Apply Newton's method: \(x_{n+1}=x_n-\frac{\frac{1}{x_n}-b}{-\frac{1}{x_n^2}} = x_n+\left(\frac{1}{x_n}-b\right)x_n^2\). 3. Simplify: \(x_{n+1}=x_n+x_n-bx_n^2=x_n(2-bx_n)\). 4. With \(b=13\) and \(x_0=0.1\), \(x_1=0.1(2-13\cdot0.1)=0.07\), \(x_2=0.07(2-13\cdot0.07)=0.0763\), and \(x_3=0.0763(2-13\cdot0.0763)=0.07691803\). 5. With \(x_0=0.2\), the iterations are \(x_1=-0.12\), \(x_2=-0.4272\), and \(x_3=-3.22689792\). Their magnitudes increase instead of approaching \(\frac{1}{13}\), so the sequence diverges. In fact, a positive initial value must satisfy \(0<x_0<\frac{2}{13}\) for this iteration to converge to \(\frac{1}{13}\).

a) \(x_{n+1}=x_n(2-bx_n)\) b) \(x_1=0.07\); \(x_2=0.0763\); \(x_3=0.07691803\) c) \(x_1=-0.12\), \(x_2=-0.4272\), and \(x_3\approx-3.2269\). The values move away from \(\frac{1}{13}\), so \(x_0=0.2\) is unsuitable.

- Use continuity and the signs of the endpoint values for part 1. - Newton's method for \(g(x)=0\) requires the first derivative of \(g\). - For part 3, treat the second derivative as the function whose zero you are finding. - Avoid rounding intermediate values too early.

1. The polynomial \(g\) is continuous. Also, \(g(2)=-2\) and \(g(3)=25\). Because the endpoint values have opposite signs, the Intermediate Value Theorem guarantees at least one zero in \([2, 3]\). 2. Differentiate: \(g'(x)=4x^3-6x^2\). Starting with \(x_0=2\), the first iteration gives \(x_1=2-\frac{-2}{8}=2.25\). The second gives \(x_2=2.25-\frac{\frac{217}{256}}{\frac{243}{16}}=\frac{8531}{3888}\approx2.1942\). 3. Let \(h(x)=g''(x)=12x^2-12x\), so \(h'(x)=24x-12\). Starting with \(x_0=2\), Newton's method gives \(x_1=\frac{4}{3}\approx1.3333\) and \(x_2=\frac{16}{15}\approx1.0667\).

1. Since \(g\) is continuous, \(g(2)<0\), and \(g(3)>0\), the Intermediate Value Theorem guarantees a zero in \([2, 3]\). 2. \(x_1=2.25\); \(x_2=\frac{8531}{3888}\approx2.1942\) 3. \(x_1=\frac{4}{3}\approx1.3333\); \(x_2=\frac{16}{15}\approx1.0667\)

- Picture the graph near a local minimum. - Compare the tangent directions on the two sides of the minimum. - Relate the derivative to the tangent slope. - Check the denominator of Newton's formula.

1. Let \(f(x)=x^3-3x+1\), so \(f'(x)=3x^2-3\). 2. At \(x_1=0.9\), \(f(0.9)=-0.971\) and \(f'(0.9)=-0.57\). Thus, \(x_2=0.9-\frac{-0.971}{-0.57}\approx-0.8035\). 3. At \(x_1=1.1\), \(f(1.1)=-0.969\) and \(f'(1.1)=0.63\). Thus, \(x_2=1.1-\frac{-0.969}{0.63}\approx2.6381\). 4. The graph has a local minimum at \(x=1\). Just left of the minimum, the tangent slope is negative, so its x-intercept lies to the left. Just right of the minimum, the slope is positive, so its x-intercept lies far to the right. 5. At \(x=1\), \(f'(1)=0\). The tangent is horizontal, so Newton's formula is undefined.

a) From \(x_1=0.9\), \(x_2\approx-0.8035\). From \(x_1=1.1\), \(x_2\approx2.6381\). b) The initial values lie on opposite sides of a local minimum, where the tangent slopes have opposite signs. c) \(f'(1)=0\), so the Newton update would divide by zero.

- Check the derivative in the denominator at \(x=2\). - Compare the derivative value at \(x=2.1\) with zero before carrying out the update. - Relate a nearly horizontal tangent to the location of its x-intercept.

1. Differentiate: \(f'(x)=3x^2-6x\). At \(x_0=2\), \(f'(2)=0\), so Newton's formula is undefined; geometrically the tangent is horizontal. 2. At \(x_0=2.1\), \(f(2.1)=-2.969\) and \(f'(2.1)=0.63\). Thus \(x_1=2.1-\frac{-2.969}{0.63}\approx6.8127\), so \(x_1\approx6.81\). 3. The graph shows that the tangent at \(x=2.1\) is nearly horizontal, so its x-intercept lies far to the right. The small denominator in Newton's update produces the same large jump algebraically.

a) \(f'(2)=0\), so Newton's update would divide by zero. b) \(x_1\approx6.81\). c) The tangent at \(x_0=2.1\) has a very small positive slope, so it reaches the x-axis only far to the right; this is the geometric reason for the overshoot.

- Use the derivative to locate horizontal tangents. - Interpret each Newton update as the x-intercept of a tangent line. - Keep fractions exact during the two iterations from \(x_0=3\). - A first step across a critical point can send later iterations to a different zero.

1. Differentiate: \(f'(x)=3x^2-6x=3x(x-2)\). The derivative is zero at \(x_0=0\) and \(x_0=2\), so the Newton update is undefined. These are the local maximum and minimum, where the graph has horizontal tangents. 2. From \(x_0=3\), \(f(3)=2\) and \(f'(3)=9\). Thus, \(x_1=3-\frac{2}{9}=\frac{25}{9}\approx2.7778\). 3. At \(x_1=\frac{25}{9}\), \(f(x_1)=\frac{208}{729}\) and \(f'(x_1)=\frac{175}{27}\). Therefore, \(x_2=\frac{25}{9}-\frac{\frac{208}{729}}{\frac{175}{27}}=\frac{12917}{4725}\approx2.7338\). 4. The values remain to the right of \(2\) and approach the right-hand zero \(1+\sqrt{3}\approx2.7321\). 5. From \(x_0=1.8\), \(x_1=1.8-\frac{-1.888}{-1.08}=\frac{7}{135}\approx0.0519\). Further iterations approach the right-hand zero. 6. From \(x_0=0.2\), \(x_1=0.2-\frac{1.888}{-1.08}=\frac{263}{135}\approx1.9481\). Further iterations approach the left-hand zero \(1-\sqrt{3}\approx-0.7321\). Therefore, neither initial value approaches the middle zero.

a) \(x_0=0\) and \(x_0=2\); the tangents are horizontal there. b) 1. \(x_1=\frac{25}{9}\approx2.7778\); \(x_2=\frac{12917}{4725}\approx2.7338\) 2. The iteration approaches \(1+\sqrt{3}\approx2.7321\). c) From \(x_0=1.8\), \(x_1=\frac{7}{135}\approx0.0519\) and the sequence approaches the right-hand zero. From \(x_0=0.2\), \(x_1=\frac{263}{135}\approx1.9481\) and the sequence approaches the left-hand zero.