y = x^3
I want x for y = 25
dy = 3 x^2 dx
start with x = 3
then y1 = 3^3 = 27
y2 - y1 = 3 * 9 (x2 - x1)
25 - 27 = 27 (x2 -3)
X2 -3 = -2/27
x2 = 2.93
do it again with x1 = 2.93 etc
right answer is
2.924017738
estimate cube root of 25 by linear approximation
f(x+delta(x))= f(x) + (df(x)/x) * deltax
3 + 1/3x^(-2/3) * -2
3 - 2/3x^(-2/3)
3 - (2/3* ((cube root(9)))
is this right? thanks
2 answers
Thank you! :-)