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"")#