Дано:

y=2/(x-3)

dy=0,075

x0=1

_________

∆x—?

Решение:

dy = ∆y / ∆x → ∆x = ∆y / dy. (1)

∆y = y1 - y0 = y(x0+∆x) - y(x0) =

= 2 / (x0+∆x-3) - 2 / (x0-3) =

Подставляем х0=1:

= 2 / (1+∆x-3) - 2 / (1-3) =

= 2 / (∆x-2) - 2 / -2 =

= 2 / (∆x-2) + 1 = 2 / (∆x-2) + (∆x-2) / (∆x-2) =

= (2+∆x-2) / (∆x-2) = ∆x / (∆x-2). (2)

Подставляя ∆у из(2) в (1) получаем:

∆x = ∆y/dy или

∆х = (∆х / (∆х-2))/dy

Преобразуем:

∆x = ∆х / (∆х-2)*dy. |:∆x

1 = 1 / (∆x-2)*dy

(∆x-2)*dy= 1 / 1

(∆x-2)*dy = 1

Подставляя dy=0,075 получаем:

(∆x-2)*0,075=1

∆x-2=1/0,075

[ 1/0,075 = 1000 / 75 = 40 / 3 = 13 1/3 ]

∆x-2=13 1/3

∆х=2+13 1/3

∆х=15 1/3