Question

Use Newton's Method to approximate the root of x lnx=2 accurate to five decimal places. Begin by sketching a graph.

this one i am having problems with

Answer 1

\[x\ln(x)=2\]
\[x\ln(x)-2=0\] make a guess

Answer 2

doesnt even have to be that good of a guess. i guess 2 and get
\[f(2)=2\ln(2)-2=-.61370...\]

Answer 3

ohhh okay i understand what to do now...i can pick any number right?

Answer 4

the derivative of
\[x\ln(x)-2\] is
\[1+\frac{1}{x}\]

Answer 5

yeah some number that will give you something close to 0. doesn't even have to be that close. i picked 2. next number will be
\[2-\frac{f(2)}{f'(2)}\]

Answer 6

so i get
\[2-\frac{-.6137}{(1+\frac{1}{-.6137})}\]

Answer 7

now i need a calculator for sure

Answer 8

gives 1.02504

Answer 9

next number is
\[1.02504-\frac{f(1.02504)}{(1+\frac{1}{1.02504})}\]

Answer 10

oh crap! this is a tricky one

Answer 11

it is bt i think they want an approximation that is close to 2

Answer 12

if you look at the graph you will see that the tangent lines head off in the wrong direction. you need to pick somehing bigger than 2, say 2.5. that should work. newton's method does not always work right and if you look at the graph you will see why. tangent lines get farther from the roots, not close. try looking here
http://www.wolframalpha.com/input/?i=y%3Dx*ln%28x%29-2