How do you use Newton's method to find the approximate solution to the equation #x+1/sqrtx=3#?
1 Answer
I got
Newton's method (or the Newton-Raphson method) entails:
#x_(n ew) = x_(old) - (f(x_(old)))/(f'(x_(old))#
and this new guess,
First, I'd move all the
#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
#4 -> 2.4 -> 2.347433739 -> 2.347296356 -> 2.347296355#
so it converges to
Check the solution:
#2.347296355 + 1/sqrt(2.347296355) stackrel(?)(=) 3#
#2.347296355 + 0.6527036447 = 3 = 3# #color(blue)(sqrt"")#