It seems that most calculus tests I receive to check my programs with, and with regards to graphing a function by hand, they always have you find: ” critical points”. Then the answer is always just what “x” equals. You factor the first derivative f ‘(x), find the value of x or x’s and mark it down on your test problem to get it correct x=2 or x=5 or whatever. If I was your professor (you wish) you’d have gotten only partial credit because you found the “critical numbers” and not the “critical points”. However, critical points are actually the value of (x,y). You find the values of x, from the first derivative, plug those values into the original function f(x) to find the value of y, and you have the critical point or points (x,y). When tests ask for critical numbers the professor actually means critical numbers. There is a difference!! In my program I find both for you critical numbers and critical points (step by step of course) and leave it up to you as to what or how your professor teaches this, in most cases teaching it incorrectly, what else is new?
Solution:Step1First replace x with (x-Δx) in the given focutinn f(x) to obtainf(x+Δx) = 2(x+Δx)+1 = 2x + 2Δx + 1 Step2Rewrite f(x-Δx) as (2x + 2Δx + 1) and f(x) as (2x + 1) into the definition of the derivative limit to obtain: lim [(2x + 2Δx + 1) – (2x+1)] / ΔxΔx-> 0 Step 3Combine like terms on the numerator and the reduced expression becomes lim [2Δx] / ΔxΔx-> 0 Step 4Cancel out Δx from both numerator and denominator and finally you get the derivative Derivative of f (x) = 2x+1 = 2 -Note there was no need to substitute Δx with 0 since it was canceled out.
considering that to solve for the x sub n+1 value you need to be able to compute f(x) for any x value, why use netwon’s method at all. With the ability to solve for f(x) for any x, one could likely end up at the zero faster from a guess and check method, especially since we already know the general shape of the graph usually. certainly solving for the zero would be much quicker. Im not trying to be a jerk, just trying to find an answer that our calc teacher cant provide us with.