(a) Use Newton's method with x1 = 1 to find the root of the equation x3 − x = 4
correct to six decimal places.
x =

(b) Solve the equation in part (a) using x1 = 0.6 as the initial approximation.
x =

(c) Solve the equation in part (a) using x1 = 0.57. (You definitely need a programmable calculator for this part.)
x =

(d) Graph
f(x) = x3 − x − 4
and its tangent lines at
x1 = 1,
0.6, and 0.57.
WebAssign Plot WebAssign Plot
WebAssign Plot WebAssign Plot

9514 1404 393

  x ≈ 1.796322 . . . in all cases

Step-by-step explanation:

For Newton's method iteration, we like to use a function such that we are finding a zero of the function. Here, this means we want to rearrange the equation to ...

  f(x) = x^3 -x -4

so that we are finding x for f(x) = 0.

The iterator is ...

  next x = x - f(x)/f'(x) . . . . . where f'(x) is the first derivative

Many graphing calculators can accept the notation f'(x), so we do not actually have to determine the derivative algebraically. In the first attachment, we have used the function f1(x) to compute the next iteration value ("next x").

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a) In the first attachment, we have used a math package to define the function, the iterator, and an accumulator of iteration values. The N[ ] function causes the display of 7 significant digits. Computations are performed to 30 significant digits. Starting values are the third argument from the right end of the expression doing the computation. They are 1, 6/10, 57/100.

It shows that it takes 6 iterations starting from x=1 to get to 6 decimal places of accuracy. The 7th iteration is computed so that we can compare the first 6 decimal places. The final iteration gives a value of x of 1.796322.

We know that Newton's method converges quadratically (from a close-enough starting point), so 4 good decimal digits in the 5th iteration can be expected to become 8 good decimal digits in the 6th iteration. That is, if two iterations agree in the first 3 decimal digits, then the second of those can generally be considered accurate to at least 6 decimal digits.

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b) Starting from x=0.6, more iterations are required. The attachment shows that 6-decimal place accuracy is achieved in 13 iterations. More iterations are required because the starting point is closer to the function turning point, so initial iterations tend to diverge from the root.

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c) Starting from x=0.57, 20 iterations are required. This starting point is on the wrong side of the turning point of the function, so the initial iteration sends the computation a significant distance in the wrong direction.

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d) The graph and tangent lines are shown in the second attachment. Successively smaller starting values have tangents with successively smaller slopes.

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation: