Free Math Worksheets

- 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.

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.

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.

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 \(x_2 = \frac{4}{3} - \frac{\frac{13}{27}}{9} = \frac{311}{243} \approx 1.2798\).

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

- First find the derivative of the function. - For the first iteration, evaluate \(f(x_0)\) and \(f'(x_0)\). - Use your value of \(x_1\) as the starting value for the second iteration. - Evaluate the function at each new approximation to check that the residual is getting closer to zero.

1. Differentiate the function: \(f'(x) = 1.5x^2 + 1\). 2. At \(x_0 = 2\), \(f(2) = 0.5 \cdot 2^3 + 2 - 4 = 2\) and \(f'(2) = 1.5 \cdot 2^2 + 1 = 7\). Thus, \(x_1 = 2 - \frac{2}{7} = \frac{12}{7} \approx 1.7143\). 3. At \(x_1 = \frac{12}{7}\), \(f\left(\frac{12}{7}\right) = \frac{80}{343}\) and \(f'\left(\frac{12}{7}\right) = \frac{265}{49}\). 4. Therefore, \(x_2 = \frac{12}{7} - \frac{\frac{80}{343}}{\frac{265}{49}} = \frac{12}{7} - \frac{16}{371} = \frac{620}{371} \approx 1.6712\).

\(x_1 \approx 1.7143\); \(x_2 \approx 1.6712\)

- 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.

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.

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.

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 \(x\approx1.637\).

- Write the equation in the form \(f(x)=0\). - Use the first derivative in Newton's formula. - Carry out the iterations in order and keep the intermediate values. - Stop when consecutive values round to the same hundredth.

1. Let \(f(x)=x^3-3x^2+1\). Then \(f'(x)=3x^2-6x\). 2. Apply \(x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\) beginning with \(x_0=0.5\). 3. The successive approximations are \(x_1=\frac{2}{3}\approx0.666667\), \(x_2\approx0.652778\), and \(x_3\approx0.652704\). 4. The last two values agree to two decimal places, so the solution is approximately \(0.65\).

\(x\approx0.65\)

- 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 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) \(x_1=4-\frac{3.5}{4}=3.125\)

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

1. At \(x_0=2.5\), \(f(2.5)=6.625\). The tangent crosses the x-axis slightly to the right of \(x=2\). 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 \(2.05\). b) \(x_1=\frac{121}{59}\approx2.051\)

- Recall the Newton iteration formula. - Identify the role of the first derivative in the formula. - Think about how a tangent line is constructed at a point on a graph. - Determine where that tangent crosses the x-axis.

1. Differentiate: \(f'(x)=0.75x^2-1\). 2. At \(x_0=-3\), \(f(-3)=-2.75\) and \(f'(-3)=5.75\). 3. Apply Newton's formula: \(x_1=-3-\frac{-2.75}{5.75}=-3+\frac{11}{23}=-\frac{58}{23}\approx-2.52\). 4. Geometrically, \(x_1\) is the x-coordinate where the tangent to the graph at \((-3, -2.75)\) intersects the x-axis.

a) \(x_1=-\frac{58}{23}\approx-2.52\) b) \(x_1\) is the x-coordinate of the x-intercept of the tangent line at \(x_0=-3\).

- 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 \(3\).

- Use the power rule to differentiate the polynomial. - Newton's update is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). - Use your value of \(x_1\) as the input for the second iteration. - In each spreadsheet formula, use the cell reference B1 in place of \(x\).

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, \(x_2 = 2.1 - \frac{0.061}{11.23} \approx 2.09457\).

1. \(f'(x) = 3x^2 - 2\) 2. C1: `=B1^3-2*B1-5`; D1: `=3*B1^2-2`; E1: `=B1-C1/D1` 3. \(x_1 = 2.1\); \(x_2 \approx 2.09457\)

- 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.

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.

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 \(x \approx 0.6180\).

- Express the slope using the first derivative. - Rewrite the slope condition as an equation equal to zero. - Use Newton's method for the resulting equation. - Keep the approximation inside the given interval.

1. Differentiate: \(f'(x)=x^3-3x^2+5\). 2. Set the slope equal to \(2\): \(x^3-3x^2+5=2\), so solve \(F(x)=x^3-3x^2+3=0\). 3. Using Newton's method with \(x_0=2.5\), \(F'(x)=3x^2-6x\). 4. The first two approximations are \(x_1=2.5-\frac{-0.125}{3.75}\approx2.5333\) and \(x_2\approx2.5321\). 5. Therefore, \(x^*\approx2.53\).

\(x^*\approx2.53\)

- 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.

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 points 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.

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.

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 \(0.0155\).

- Use \(1\,\text{L}=1\,\text{dm}^3\). - Rewrite the volume condition as an equation of the form \(f(h)=0\). - Apply Newton's update to the current value of \(h\). - Evaluate both the function and its first derivative at each approximation.

1. Since \(120\,\text{L}=120\,\text{dm}^3\), solve \(f(h)=0.5h^3+9\pi h-120=0\). 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 \(f'(3)=13.5+9\pi\approx41.774\). Thus, \(h_1=3-\frac{f(3)}{f'(3)}\approx3.518907\), so \(h_1\approx3.519\). 4. Using the unrounded value of \(h_1\), \(h_2=h_1-\frac{f(h_1)}{f'(h_1)}\approx3.491552\), so \(h_2\approx3.492\). 5. The required water depth is approximately \(3.492\,\text{dm}\).

\(h_1\approx3.519\) \(h_2\approx3.492\) The required water depth is approximately \(3.492\,\text{dm}\).

- 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 \(x>0\).

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.

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 signs at the interval endpoints. - Apply Newton's formula one step at a time. - Sketch the tangent lines at the repeated values. - Consider what happens when an approximation repeats.

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 \(x_1=1\), \(x_2=1-\frac{1}{1}=0\). 5. From \(x_2=0\), \(x_3=1\). 6. The iteration alternates between \(0\) and \(1\). The tangent at \(x=0\) meets the x-axis at \(x=1\), and the tangent at \(x=1\) meets it at \(x=0\). Newton's method is trapped in a two-cycle instead of approaching the zero near \(-1.7693\).

a) \(f(-2)=-2\) and \(f(-1)=3\), so a zero lies in \([-2, -1]\). b) \(x_1=1\); \(x_2=0\); \(x_3=1\) c) The sequence cycles between \(0\) and \(1\), so it does not converge to the actual zero.

- 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.

- 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) \(x_2=1.2\)

- 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 \(2.93\).

- 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 point 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, \(x\approx0.125\) is unsuitable because it is a visible local maximum and \(f'(0.125)\approx0\).

a) \(x_1\approx2.694\); \(x_2\approx2.586\) b) One unsuitable initial value is \(x\approx0.125\). The tangent is horizontal there, so Newton's formula would divide by zero.

- 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. - Check the end behavior of the first derivative. - Apply Newton's method to the first derivative, because a critical point occurs where \(f'(x)=0\).

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 \(f''(x)\geq3>0\) 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. Apply Newton's method to \(f'(x)=0\): \(x_{n+1}=x_n-\frac{f'(x_n)}{f''(x_n)}\). 4. 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 function \(f'\) is strictly increasing from \(-\infty\) to \(\infty\), so it has exactly one zero. 2. The unique critical point occurs at \(x\approx-1.300\), and it 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.

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 denominator of Newton's formula at \(x=2\). - Relate a zero derivative to a horizontal tangent. - Calculate the function and derivative values separately before updating. - Consider where an almost horizontal tangent meets the x-axis.

1. Differentiate: \(f'(x)=3x^2-6x\). 2. At \(x_0=2\), \(f'(2)=0\), so Newton's formula is undefined. Geometrically, the tangent at the local minimum is horizontal. 3. At \(x_0=2.1\), \(f(2.1)=-2.969\) and \(f'(2.1)=0.63\). Therefore, \(x_1=2.1-\frac{-2.969}{0.63}\approx6.8127\), so \(x_1\approx6.81\). 4. The small tangent slope near the local minimum produces an x-intercept far from the starting point and from every zero. This is an example of Newton's method overshooting when the derivative is close to zero.

a) \(f'(2)=0\), so the Newton update would divide by zero. b) \(x_1\approx6.81\) c) The first approximation lies far beyond all three zeros because the tangent slope at \(x_0=2.1\) is very small.

- 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.

- 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 point.

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 \(x_1=3\).

- 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.