Root approximation (1 Viewer)

[wagig]

Question:
x = 0.5 is an approximate solution to e^x = 1/x. Using one applications of newton's method, find another approximation.

Is my answer valid? I'm not sure if the f(x) i've chosen is the correct one to use.
f(x) = -1 + xe^x
f'(x) = xe^x + e^x
therefore by newton's method, x is approx. equal to 0.5 - f(0.5)/f'(0.5) which equals 0.571 (3 D.P).

 

[seventhroot]

hehe root


yea that looks right (assuming you subbed it into the calc correctly)

 

[acronical]

It seems ok in this case but remember that if the function is not continuous or is not differentiable for all x in given domain and codomain then you can't multiply it by x. However for HSC and this particular question it seems fine as you aren't going near zero where this particular function is undefined. Using f(x) = e^x - 1/x I got 0.5621873009 and using a computer software it solved it as 0.5671432904. Luckily this time your solution was the better approximation.