The function f(x) = x^3 - x^2 - 3x + 15 has been graphed. Starting with the initial approximation x_0 = -2, we want to use two applications of Newton's method to find a better approximation to the single x-intercept.
Evaluate f(x_0)
Evaluate f'(x_0)
Starting with x_0=-2, use a second application of Newton's method to find a better approximation, x_1, correct to 2 decimal places if necessary.
Starting with x_1=-2.69, use a second application of Newton's method to find a better approximation, x_2, correct to 2 decimal places if necessary.
Why would x_0=0 not be a good initial approximation for the x-intercept?
For the following functions with their initial approximation:
Find the exact value of f(x_0).
Find the exact value of f'(x_0).
Starting with an initial approximation of x_0, use one application of Newton's method to find a better approximation, x_1. Give your answer as an exact value.
f(x) = 3 x + 7 \ln x,\, x_0 = 1
f(x) = 2 \tan x + \ln x,\, x_0 = 2
Consider the equation x^{2} - 13 = 0. Starting with the initial approximation x_0 = 3.8, we want to use one application of Newton's method to find a better approximation to the positive root of the equation.
Newton's method of approximating roots states: x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. To use this method to find a better approximation to the solution of the x^{2} - 13 = 0, what must be f(x)?
Evaluate f(x_0).
Evaluate f'(x_0).
Starting with an initial approximation of x_0=3.8, use one application of Newton's method to find a better approximation, x_1, correct to 2 decimal places if necessary.
Consider the equation x^{3} + x - 4 = 0. Starting with the initial approximation x_0 = 1, we want to use one application of Newton's method to find a better approximation to the positive root of the equation.
Newton's method of approximating roots states: x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. To use this method to find a better approximation to the solution of the x^{3} + x - 4 = 0, what must be f(x)?
Evaluate f(x_0).
Evaluate f'(x_0).
Starting with an initial approximation of x_0=1, use one application of Newton's method to find a better approximation, x_1, correct to 2 decimal places if necessary.
Consider the equation x^5-3x+5 = 0. Starting with the initial approximation x_0 = \frac{-3}{2}, we want to use one application of Newton's method to find a better approximation to the root of the equation.
Newton's method of approximating roots states: x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. To use this method to find a better approximation to the solution of x^5-3x+5 = 0, what must be f(x)?
Find the exact value of f(x_0)
Evaluate f'(x_0)
Starting with an initial approximation of x_0 = \frac{-3}{2}, use one application of Newton's method to find a better approximation, x_1, correct to 2 decimal places if necessary.
Here is a graph of f(x).
Fill in the gaps to complete the following statement that explains Newton's method.
Geometrically, Newton's method starts with the initial approximation of x=\frac{-3}{2}, and finds the gradient of the tangent to the curve at that point. This is what f'\left(\frac{-3}{2}\right) represents. Knowing the gradient of the tangent and a point \left(\frac{-3}{2},\,f\left(\frac{-3}{2}\right)\right) on the tangent line, it then finds the equation of the tangent line. Finally it looks for the x-intercept of the tangent line, which gives us our next best approximation x_1=-1.59.
If we used x_0=\pm\sqrt[⬚]{⬚} as the initial approximation, we would get f'(x_0)=⬚. The tangent would be parallel to the x-axis, and so it would not cut the x-axis. Therefore x_0 = \pm\sqrt[⬚]{⬚} would not be a good initial approximation of the real root.
Consider the equation \frac{x^{2} - 3 x + 1}{x^{2} + 1} = 0. Starting with the initial approximation x_0 = 0.5, we want to use one application of Newton's method to find a better approximation to the root of the equation.
Newton's method of approximating roots states: x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. To use this method to find a better approximation to the solution of \frac{x^{2} - 3 x + 1}{x^{2} + 1} = 0, what must be f(x)?
Find the exact value of f(x_0)
Evaluate f'(x_0)
Starting with an initial approximation of x_0 = 0.5, use one application of Newton's method to find a better approximation, x_1, correct to 2 decimal places if necessary.
Complete the following statement:
We can't use x_0 = 1 as our initial approximation, because ⬚ = 0.
When a car engine is turned off, its temperature is 71 \degree C. It is then left outside in the ( - 11 ) \degree C cold, and t minutes after being turned off, the engine's temperature T is given by T(t) = 82 e^{ - 0.03 t} - 11.
After approximately t_0 = 69 minutes, the engine's temperature reaches 0 \degree C. We want to use Newton's method to find a better approximation for this time.
What equation are we trying to find a better approximation for? Select all the correct options.
Evaluate T(t_0), rounding to 2 decimal places.
Evaluate T'(t_0), rounding to 2 decimal places.
Starting with an initial approximation of t_0 = 69 minutes, and using the rounded values of the previous parts, use one application of Newton's method to find a better approximation, t_1, to the number of minutes it takes for the car engine to reach 0 \degree C. Round your answer to 2 decimal places if necessary.
Consider the functions h\left(x\right)=3\ln x and g\left(x\right)=x. We want to use Newton's method to find an approximation of their point of intersection.
Which function would allow us to use Newton's method to do this?
Consider the given table of values:
| x | 1 | 1.5 | 2 | 2.5 |
|---|---|---|---|---|
| f(x) | -1 | -0.28 | 0.08 | 0.25 |
What is the narrowest interval of x in which the x-intercept of f(x) lies?
Between x=⬚ and x=⬚
Starting with x_0=1.5, we will now use one application of Newton's method to find a better approximation to the point of intersection of h(x) and g(x).
First, find the exact value of f(x_0)
Find the exact value of f'(x_0)
Starting with an initial approximation of x_0=1.5, use one application of Newton's method to find a better approximation, x_1, to the point of intersection of h(x) and g(x). Round your answer to 2 decimal places if necessary.
Consider the equation \cos x+\sin x=2x. Starting with the initial approximation x_0=0.6, we want to use one application of Newton's method to find a better approximation to the root of the equation.
Newton's method of approximating roots states: x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. To use this method to find a better approximation to the solution of \cos x+\sin x=2x, what must be f(x)?
Find the exact value of f(x_0)
Find the exact value of f'(x_0)
Starting with an initial approximation of x_0 = 0.6, use one application of Newton's method to find a better approximation, x_1, correct to 2 decimal places if necessary.
Consider Newton's method of approximating roots: x_n=x_{n-1}-\frac{f\left(x_{n-1}\right)}{f'\left(x_{n-1}\right)}. If Newton's method were applied a second time to find a better approximation than x=0.71, what values would need to be substituted into the formula to find x_2?