How do you use Newton's method to find the approximate solution to the equation `x+1/sqrtx=3`?

I got `x = 2.347296355`.

Newton's method (or the Newton-Raphson method) entails:

`x_(n ew) = x_(old) - (f(x_(old)))/(f'(x_(old))`

and this new guess, `x_(n ew)`, approaches the true answer from above, while `x_(old)` is any guess in the reals. Depending on the guess, it may not converge, but pick something reasonable.

First, I'd move all the `x` and constant terms to the left:

`x + 1/sqrtx - 3 = 0`

Then, taking the derivative gives the general form:

`f'(x) = 1 + (-1/2x^(-"3/2")) = 1 - 1/(2x^"3/2")`

So, if you write this in your TI-83+ calculator:

`"[insert guess]" -> x`

`x - "("x + 1"/"sqrtx - 3")/("1 - 1"/("2x"^"("3/2")"))" -> x`

which actually looks like this:

`[x_(old) - (x_(old) + 1/sqrt(x_(old)) - 3)/(1 - 1/(2(x_(old))^("3/2")))] -> x_(n ew)`

and then press enter several times, you should get the following output for an initial guess of `x = 4`:

`4 -> 2.4 -> 2.347433739 -> 2.347296356 -> 2.347296355`

so it converges to `color(blue)(2.347296355)`.

Check the solution:

`2.347296355 + 1/sqrt(2.347296355) stackrel(?)(=) 3`

`2.347296355 + 0.6527036447 = 3 = 3` `color(blue)(sqrt"")`