Consider the following equations. f(x)= x^3 +3x^2 -2x+1 g(x)= x^2-5x+4 approximate the solution to the equation f(x) = g(x) using three iterations of successive approximation.

a. x =11/16

b. x=13/16

c. x=3/8

d. x= 7/8

A. x = 11/16

Step-by-step explanation:

For the purpose here, it is convenient to rearrange the equation to f(x) -g(x) = 0. We know the root will be in the interval [0, 1] because (f-g)(0) = -3 and (f-g)(1) = +3. At each iteration, we evaluate (f-g)(x) at the midpoint of the interval to see which of the interval end points can be moved and still bracket the root.

Using the bisection method starting with the interval [0, 1] we find f(1/2)-g(1/2) < 0, so we can move the interval limits to [1/2, 1].

For the next iteration, we find f(3/4) -g(3/4) > 0, so we can move the interval limits to [1/2, 3/4].

For the third iteration, we find f(5/8) -g(5/8) < 0, so we can move the interval limits to [5/8, 3/4].

Then the root is approximately the middle of that interval:

x ≈ (5/8 +3/4)/2 = 11/16

_____

This value of x is 0.6875. The root is closer to 0.639802004233. The bisection method takes about 3 iterations for each decimal place of accuracy. Other methods can nearly double the number of accurate decimal places on each iteration.

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

probably the last option

step-by-step explanation: