Question

Approximate Solutions of Equations Question?

Show that the equation `x^3-12x+10=0` has three real roots and determine two consecutive numbers between which each of the roots lie. Hence, by applying the half-interval method, approximate the root nearest to the origin to 1dp.

Thanks!

Answer 1

The root near `0` is approximately `0.9`

Explanation:

Let:

`f(x) = x^3-12x+10`

Then:

`f(-4) = -64+48+10 = -6 < 0`

`f(-3) = -27+36+10 = 19 > 0`

`f(0) = 0-0+10 = 10 > 0`

`f(1) = 1-12+10 = -1 < 0`

`f(2) = 8-24+10 = -6 < 0`

`f(3) = 27-36+10 = 1 > 0`

So `f(x)` has real zeros in `(-4, -3)`, `(0, 1)` and `(2, 3)`

The root nearest the origin is the one in `(0, 1)`

Personally, I would linearly interpolate to find approximate root `10/11 ~~ 0.9`, but let's bisect the interval to find the approximation...

`f(1/2) = (1/2)^3-12(1/2)+10 = 1/8-6+10 = 33/8 > 0`

`f(3/4) = (3/4)^3-12(3/4)+10 = 27/64-9+10 = 91/64 > 0`

`f(7/8) = (7/8)^3-12(7/8)+10 = 343/512-21/2+10 = 87/512 > 0`

`f(15/16) = (15/16)^3-12(15/16)+10 = 3375/4096-45/4+10 = -1745/4096 < 0`

So the zero lies between `7/8 = 0.875` and `15/16 = 0.9375`, so to one decimal place is `0.9`.

Bonus

A more efficient way of finding the zeros is Newton's method.

Given:

`f(x) = x^3-12x+10`

the derivative of `f(x)` is:

`f'(x) = 3x^2-12`

Then given an approximate zero `x=a`, a better approximation is:

`a-(f(a))/(f'(a)) = a-(x^3-12x+10)/(3x^2-12)`

In our example, we could begin by with `a=1`.

Then a better approximation is:

`1-(f(1))/(f'(1)) = 1-(1-12+10)/(3-12) = 1-(-1)/(-9) = 8/9 = 0.bar(8) ~~ 0.9`

We can repeat to get a better approximation:

`8/9-(f(8/9))/(f'(8/9)) = 8/9-(512/729-32/3+10)/(64/27-12)`

`color(white)(8/9-(f(8/9))/(f'(8/9))) = 8/9-(26/729)/(-260/27)`

`color(white)(8/9-(f(8/9))/(f'(8/9))) = 8/9+1/270`

`color(white)(8/9-(f(8/9))/(f'(8/9))) = 241/270`

`color(white)(8/9-(f(8/9))/(f'(8/9))) = 0.8bar(925) ~~ 0.8926`