Newton’s method practice question 4
The equation has exactly one real solution. Between which two consecutive whole numbers does it lie, and which of the two is the better starting point for Newton’s method?
Put every term on one side, , and evaluate at . Where does the sign change?
A continuous that is negative at one whole number and positive at the next has a zero between them. Which end has closer to 0?
Theory UBC Math 100 §10.1Key results
Worked solution
Strategy
Write the equation as f(x) = 0 and evaluate f at whole numbers until the sign changes: the solution lies between those two. The end where f is much nearer 0 is usually the nearer one, and makes the better start.
Working
f is increasing everywhere, so it has exactly one zero, as the question says.
f(1) and f(2) are both negative, so no sign change shows between 1 and 2. f(2) < 0 < f(3), so the zero lies between 2 and 3.
From 2 to 3, f climbs 23, and the zero is where it has climbed only 1, roughly a 23rd of the way along. So the solution is just above 2, and x0 = 2 is the start to take.
One step from x0 = 2 already lands within 0.002 of the solution, 2.0611 to four places.
Answer
The answer is Between 2 and 3, and the better start is x₀ = 2, where f is −1..
Check
f(2.06) = 8.742 + 8.24 − 17 ≈ −0.02 and f(2.07) ≈ +0.15, so the solution is between 2.06 and 2.07, close to 2.
Where this goes wrong
Choosing an interval because f is small at one end, without a sign change. Between 1 and 2 the function is negative at both ends, so nothing places a zero there; the sign changes between 2 and 3.
The idea behind it
The Newton’s method lesson, a figure to move, in a few steps
Leans on
The Intermediate Value Theorem, from Continuity
Try a similar question
Let . Starting from , use one step of Newton’s method to find .Saved in this browser only.