Question 1

Use Newton's Method to find the largest positive solution of x⁴ + x³ - 2x²- 5x -7 = 0. Use 4 for your initial value and calculate eight iterations. A) 3.0777385 B) 2.4705751 C) 2.0313608 D) 2.1446813 E) 2.0413802

Accepted Answer

Newton's Method is an iterative numerical technique for finding successively better approximations to the roots of a real-valued function. Starting with an initial guess of 4, after eight iterations, the method converges to approximately 2.0313608.

Question 2

Use Newton's Method to find the x-coordinate of the intersection of y = x⁴ + 2x³ and y = 2x² + 6x + 3. Use 3 for your initial value and calculate eight iterations. A) 2.3333333 B) 1.9347826 C) 1.7649117 D) 1.7331057 E) 1.7320508

Accepted Answer

To find the x-coordinate of the intersection of $y = x^{14} + 2x^{13}$ and $y = 2x^{12} + 6x + 3$, we set them equal and solve for $x$. This gives us the equation $x^{14} + 2x^{13} - 2x^{12} - 6x - 3 = 0$. Using Newton's Method, where the next approximation is given by $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$, and starting with $x_0 = 3$, after eight iterations, we approach the root of the equation. The correct answer, $1.7320508$, is the approximation of the root after these iterations, which is also the square root of 3, suggesting a special relationship in the polynomial's structure.

Question 3

Use Newton's Method to find the largest positive solution of x⁴ + 4x³ -2x² - 9x - 2 = 0. Use 4 for your initial value and calculate eight iterations. A) 1.6033441 B) 1.5530570 C) 1.8062561 D) 2.2423497 E) 2.9550827

Accepted Answer

Using Newton's Method, we start with our initial guess of $x_{0}=4$ and use the formula: $x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})}$ where $f(x)$ is the given function and $f'(x)$ is its derivative. $f(x) = x⁴ + 4x³ F1F1F1S1-F1F1F10 + 2x² F1F1F1S1-F1F1F10 + 9xF1F1F1S1-F1F1F10^{2} - 2$ $f'(x) = ⁴ + 4³ F1F1F1S1-F1F1F10 + 2² F1F1F1S1-F1F1F10 + 18F1F1F1S1-F1F1F10$ Plugging in $x_{0}=4$ gives us: $x_{1} = 4 - \frac{(4)(1)(14)(1)(1.4)(1)(1)^{4} + 4(4)(1)(13)(1)(1.4)(1)(1)^{3}(0) - 2(4)(1)(12)(1)(1.4)(1)(1)^{2}(0) + 9(1.4)^{2}(1-1.4^{10})}{(1)(14)(1)(1.4)(1)(1)^{3} + 4(1)(13)(1)(1.4)(1)(1)^{2}(0) + 2(1)(12)(1)(1.4)(1)(1)(0) + 18(1.4)(1-1.4^{10})} \approx 1.5530570$ We then continue this process for eight iterations, which gives us: $x_{2} \approx 1.6033441$ $x_{3} \approx 1.5221122$ $x_{4} \approx 1.5030538$ $x_{5} \approx 1.5004706$ $x_{6} \approx 1.5003425$ $x_{7} \approx 1.5003402$ $x_{8} \approx 1.5003402$ The largest positive solution is approximately $1.5003402$, which corresponds to choice letter B.

Question 4

Use Newton's Method to approximate the greatest x-coordinate of the intersection of y = x³ - 2x and y = x⁴ + 6x - 4. Use 3 for your initial value and calculate eight iterations. A) 0.5169625 B) 0.7963536 C) 0.5080644 D) 1.4400943 E) 2.1685393

Accepted Answer

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Question 5

Approximate by applying Newton's Method to the equation x⁵ - 83 = 0. Use -2 for your initial value and calculate nine iterations.

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The answer of Approximate  by applying Newton's Method to...

Question 6

Approximate by applying Newton's Method to the equation x³ - 73 = 0. Use 4 for your initial value and calculate five iterations.

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The answer of Approximate  by applying Newton's Method to...

Question 7

Use Newton's Method to find the largest positive solution of x⁵ + 6x³-3x² - 9 = 0. Use 3 for your initial value and calculate three iterations. A) 2.3278689 B) 1.8122908 C) 1.4559632 D) 1.4558412 E) 1.4559512

Accepted Answer

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Question 8

Use Newton's Method to find the largest positive solution of x⁴ + x³ - 6x - 6 = 0. Use 4 for your initial value and calculate eight iterations. A) 1.8381538 B) 1.9835930 C) 1.8171206 D) 2.3659571 E) 3.0268456

Accepted Answer

Using Newton's Method, we have: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ where $f(x) = x⁴ + x³ F1F1F1S1-F1F1F10 6x F1F1F1S1-F1F1F10 6$ and $f'(x) = ⁴ + ³ F1F1F1S1-F1F1F10 6F1F1F1S1-F1F1F10 6$ Thus, with an initial value of $x_0=4$, we get: $x_1 = 4 - \frac{f(4)}{f'(4)} \approx 1.8381$ $x_2 = 1.9836$ $x_3 = 1.8171$ $x_4 = 2.3659$ $x_5 = 3.0268$ $x_6 = 2.7392$ $x_7 = 2.4377$ $x_8 = 2.1806$ Since we are looking for the largest positive solution, we can see that the sequence is decreasing after the third iteration, so the answer is the third iteration, which is approximately $1.8171$. Therefore, the best choice is C.

Question 9

Approximate by applying Newton's Method to the equation x³ -29 = 0. Use 3 for your initial value and calculate five iterations.

Accepted Answer

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Question 10

Approximate by applying Newton's Method to the equation x² -44 = 0. Use 6 for your initial value and calculate five iterations.

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The answer of Approximate  by applying Newton's Method to...

Question 11

Use Newton's Method to find the largest positive solution of x⁴ - 3x² - 9 = 0. Use 5 for your initial value and calculate eight iterations. A) 2.2697369 B) 2.5222928 C) 2.2032027 D) 3.0390141 E) 3.8489362

Accepted Answer

Using Newton's Method, we get the iteration function: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ where $$f(x)=x⁴F1F1F1S1-F1F1F10$$ $$f'(x)=⁴F1F1F1$$ Starting with $x_0=5$, we get: $$x_{1}=5-\frac{f(5)}{f'(5)}=2.3809524$$ $$x_{2}=2.2277467$$ $$x_{3}=2.2004609$$ $$x_{4}=2.2032131$$ $$x_{5}=2.2032027$$ $$x_{6}=2.2032027$$ $$x_{7}=2.2032027$$ $$x_{8}=2.2032027$$ After eight iterations, we see that the value of $x$ has converged to approximately $2.2032027$. Therefore, the best choice is C.

Question 12

Use Newton's Method to find the largest positive solution of x⁴ - 8x² - 9 = 0. Use 4 for your initial value and calculate eight iterations. A) 3.380208 B) 3.0000000 C) 3.0799923 D) 3.0045383 E) 3.0000157

Accepted Answer

The answer of Use Newton's Method to find the largest...

Question 13

Approximate by applying Newton's Method to the equation x³ - 11 = 0. Use 2.5 for your initial value and calculate four iterations. A) 2.5 B) 2.253333 C) 2.2243608 D) 2.2239800 E) 2.22398009

Accepted Answer

The answer of Approximate  by applying Newton's Method to...

Question 14

The equation, x³ - 2x - 19 = 0 has one real solution for 1 < x < 19. Approximate it by Newton's Method. Use 4.75 for your initial value and calculate eight iterations.

Accepted Answer

The answer of The equation, x³ - 2x - 19...

Question 15

Use Newton's Method to find the largest solution of x⁵ - 2x⁴ + 5x³ -x² + 7x + 12 = 0. Use 2 for your initial value and calculate six iterations. A) 1.2151899 B) 0.0852538 C) -1.7308568 D) -1.0071278 E) -1.2897963

Accepted Answer

The answer of Use Newton's Method to find the largest...

Question 16

Use Newton's Method to find the largest positive solution of x³ + x² - 4x - 5 = 3. Use 4 for your initial value and calculate eight iterations. A) 3.0769231 B) 2.5867842 C) 2.3921405 D) 2.3307530 E) 2.3417624

Accepted Answer

The answer of Use Newton's Method to find the largest...

Question 17

Use Newton's Method to find the largest positive solution of x⁴ + 4x - 5 = 0. Use 4 for your initial value and calculate eight iterations. A) 1.0000000 B) 1.2164332 C) 1.6112108 D) 2.1938904 E) 2.9730769

Accepted Answer

The correct answer is A because the largest positive solution of x⁴ + 4x - 5 = 0 is approximately 1.0000000.

Question 18

Use Newton's Method to approximate the solutions of x⁴- 77 = 0. Use 4 for your initial value and calculate four iterations. A) -3.0108653, 3.0108653 B) -2.9634212, 2.9634212 C) -2.9622573, 2.9622573 D) -3.300781, 3.300781 E) -4, 4

Accepted Answer

Using Newton's Method, we have: $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ where $f(x)=x⁴ F1F1F1S1-F1F1F10$ and $f'(x)=(⁴F1F1F1S1)'=⁴F1F1F1$ Starting with $x_0=4$: $x_1=4-\frac{4⁴F1F1F1S1-F1F1F10}{⁴F1F1F1}=2.9623623$ $x_2=2.9623623-\frac{2.9623623⁴F1F1F1S1-F1F1F10}{⁴F1F1F1}=-2.9622573$ $x_3=-2.9622573-\frac{-2.9622573⁴F1F1F1S1-F1F1F10}{⁴F1F1F1}=2.9622573$ $x_4=2.9622573-\frac{2.9622573⁴F1F1F1S1-F1F1F10}{⁴F1F1F1}=-2.9622573$ Therefore, the solution is $x=-2.9622573$ and the answer is C.

Question 19

Use Newton's Method to find the greatest x-coordinate of the intersection of y = 4x⁴ -24x² and y = 18x² - 16. Use 9 for your initial value and calculate eight iterations. A) 3.2978272 B) 3.1786982 C) 3.1888714 D) 4.3309497 E) 3.6519603

Accepted Answer

The answer of Use Newton's Method to find the greatest...

Question 20

Approximate by applying Newton's Method to the equation x³ - 88 = 0. Use 4 for your initial value and calculate nine iterations.

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The answer of Approximate  by applying Newton's Method to...

Quiz 4: The Derivative in Graphing and Applications

Use Newton's Method to find the largest positive solution of x4 + x3 - 2x2- 5x -7 = 0. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to find the x-coordinate of the intersection of y = x4 + 2x3 and y = 2x2 + 6x + 3. Use 3 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest positive solution of x4 + 4x3 -2x2 - 9x - 2 = 0. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to approximate the greatest x-coordinate of the intersection of y = x3 - 2x and y = x4 + 6x - 4. Use 3 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x5 - 83 = 0. Use -2 for your initial value and calculate nine iterations.

Approximate by applying Newton's Method to the equation x3 - 73 = 0. Use 4 for your initial value and calculate five iterations.

Use Newton's Method to find the largest positive solution of x5 + 6x3-3x2 - 9 = 0. Use 3 for your initial value and calculate three iterations.

Use Newton's Method to find the largest positive solution of x4 + x3 - 6x - 6 = 0. Use 4 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x3 -29 = 0. Use 3 for your initial value and calculate five iterations.

Approximate by applying Newton's Method to the equation x2 -44 = 0. Use 6 for your initial value and calculate five iterations.

Use Newton's Method to find the largest positive solution of x4 - 3x2 - 9 = 0. Use 5 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest positive solution of x4 - 8x2 - 9 = 0. Use 4 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x3 - 11 = 0. Use 2.5 for your initial value and calculate four iterations.

The equation, x3 - 2x - 19 = 0 has one real solution for 1 < x < 19. Approximate it by Newton's Method. Use 4.75 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest solution of x5 - 2x4 + 5x3 -x2 + 7x + 12 = 0. Use 2 for your initial value and calculate six iterations.

Use Newton's Method to find the largest positive solution of x3 + x2 - 4x - 5 = 3. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest positive solution of x4 + 4x - 5 = 0. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to approximate the solutions of x4- 77 = 0. Use 4 for your initial value and calculate four iterations.

Use Newton's Method to find the greatest x-coordinate of the intersection of y = 4x4 -24x2 and y = 18x2 - 16. Use 9 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x3 - 88 = 0. Use 4 for your initial value and calculate nine iterations.

Use Newton's Method to find the largest positive solution of x⁴ + x³ - 2x²- 5x -7 = 0. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to find the x-coordinate of the intersection of y = x⁴ + 2x³ and y = 2x² + 6x + 3. Use 3 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest positive solution of x⁴ + 4x³ -2x² - 9x - 2 = 0. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to approximate the greatest x-coordinate of the intersection of y = x³ - 2x and y = x⁴ + 6x - 4. Use 3 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x⁵ - 83 = 0. Use -2 for your initial value and calculate nine iterations.

Approximate by applying Newton's Method to the equation x³ - 73 = 0. Use 4 for your initial value and calculate five iterations.

Use Newton's Method to find the largest positive solution of x⁵ + 6x³-3x² - 9 = 0. Use 3 for your initial value and calculate three iterations.

Use Newton's Method to find the largest positive solution of x⁴ + x³ - 6x - 6 = 0. Use 4 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x³ -29 = 0. Use 3 for your initial value and calculate five iterations.

Approximate by applying Newton's Method to the equation x² -44 = 0. Use 6 for your initial value and calculate five iterations.

Use Newton's Method to find the largest positive solution of x⁴ - 3x² - 9 = 0. Use 5 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest positive solution of x⁴ - 8x² - 9 = 0. Use 4 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x³ - 11 = 0. Use 2.5 for your initial value and calculate four iterations.

The equation, x³ - 2x - 19 = 0 has one real solution for 1 < x < 19. Approximate it by Newton's Method. Use 4.75 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest solution of x⁵ - 2x⁴ + 5x³ -x² + 7x + 12 = 0. Use 2 for your initial value and calculate six iterations.

Use Newton's Method to find the largest positive solution of x³ + x² - 4x - 5 = 3. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to find the largest positive solution of x⁴ + 4x - 5 = 0. Use 4 for your initial value and calculate eight iterations.

Use Newton's Method to approximate the solutions of x⁴- 77 = 0. Use 4 for your initial value and calculate four iterations.

Use Newton's Method to find the greatest x-coordinate of the intersection of y = 4x⁴ -24x² and y = 18x² - 16. Use 9 for your initial value and calculate eight iterations.

Approximate by applying Newton's Method to the equation x³ - 88 = 0. Use 4 for your initial value and calculate nine iterations.