How do you use Newton's method to find the approximate solution to the equation #x+sqrtx=1#?

Question

Luke Alvarez · Accepted Answer

The solution is #x=0.3820# to 4dp.

We want to solve:

# x+sqrt(x) = 1 => x+sqrt(x) -1 =0 #

Let #f(x) = x+sqrt(x) -1# Then our aim is to solve #f(x)=0#.

First let us look at the graphs:
graph{x+sqrt(x) -1 [-2, 3, -2, 2]}

We can see there is one solution in the interval #0 le x le 1#, so will use #x=1# as our initial approximation

To find the solution numerically, using Newton-Rhapson method we will need the derivative #f'(x)#.

# \ \ \ \ \ \ \f(x) = x+sqrt(x) -1 #
# :. f'(x) = 1+1/2x^(-1/2) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \= 1+1/(2sqrt(x)) #

The Newton-Rhapson method uses the following iterative sequence

# { (x_1,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :} #

Then using excel working to 8dp we can tabulate the iterations as follows:

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is #x=0.3820# to 4dp.

Luke Alvarez · Answer

undefined To use Newton's method to find the approximate solution to the equation $x + \sqrt{x} = 1$, follow these steps:

1. Start with an initial guess for the solution, denoted as $x_0$.
2. Iterate using the formula: 
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
where $f(x) = x + \sqrt{x} - 1$ is the function representing the equation and $f'(x)$ is its derivative.
3. Repeat the iteration until the desired level of accuracy is achieved.

The derivative of $f(x) = x + \sqrt{x} - 1$ is $f'(x) = 1 + \frac{1}{2\sqrt{x}}$.