Please don't post the same problem multiple times. It only irritates people. Only volunteers work here. Paid providers sometimes respond more quickly. sometimes not.
A derivative will not give you a solution. It can be used to help you find a solution, q.v. Newton's Method.
The interpolation concept is merely keeping the solution between two existing values. There are various ways to select the next value.
I'll think you through it.
Suppose f(x) = {[(1+x)^10] - 1} - 12x
f(0) = 0
I'm guessing we're not looking for that solution.
Knowing where to start is another problem. Various equations may have several solutions, but only some solutions may make sense. For example, your problem may have something to do with calculating an internal rate of returm. Normally, interest rates don't wander too far from 10%. Start there.
f(0.10) = 0.393742
f(0.20) = 2.791736
That seems like the wrong way.
f(0.05) = 0.0288946
That's better, but it's still positive.
f(0.04) = 0.000244285
Well, that was closer than hoped. It may be just luck.
f(0.03) = -0.0160836
Aha! We finally switch signs. There MUST be a solution in (0.03,0.04). It was so close to 0.04, let's decrease the step size a bit.
f(0.039) = -0.001927405
Still negative.
f(0.0399) = 0.000021589
Back to positive. Now we know it's here (0.039,0.0399), and it is likely to be substantially closer to 0.0399. Chage the step size again.
f(0.0398) = -0.000199877
Negative. Changing signs is good. We are on the right track.
f(0.03985) = -0.000089298
That gave us another decimal place, but it's still negative.
f(0.03987) = -0.00004498
Too conservative. It didn't change signs.
f(0.03989) = -0.000000613
Now we're getting somewhere. Still negative!
f(0.039891) = 0.000001607
Whoa! That was way too big a step.
f(0.0398901) = -0.000000391
One more and I think I've maxed out my calculator.
f(0.0398902) = -0.000000169
I wasn't quite as impressed by that as I had hoped. One more should do it.
f(0.0398903) = -0.000000053
You get the idea. Keep going in one direction until it changes sign or otherwise tells you you are going the wrong way. You can use some judgment on what value to pick next or just define a simple algorithm. Bisection is popular.
