Approximate Solutions of Equations Question?

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#