2 Types Of Gradient

Introduction: The Aim of this portfolio is to Determine by two methods the multinomial Gradient Functions. Previously we have been taught that the fixed side of a straight line that connects two separate points is given by the gradient formula: m = (y2 y1)÷(x2 x1) The problem arises in the level that in a thin outd line (e.g. y = x2) the gradient of the ignore constantly changes at different points along the curve. and then(prenominal) this indicates that instead of being a fixed formula, the gradient of a curve has a function and not a constant. In this portfolio, I plan to compare the Function for the gradient with the conventional tangent method, and to see if these are consistent with each differently in that they will refer the same result when take to on identical charts. Method 1: Chord Gradients: enforce the represent shown in the Assignment take sheet, we see that for the graph of the function y = f (x) two points on the curve are labeled P and Q, the coordinates of P = (x, f(x)), and the coordinates of Q = (x+h, f(x+h)) The definition of the check gradient function g (x,h) of the line PQ is simply the unskilful gradient formula stated earlier where m = (y2 y1)÷(x2 x1), aft(prenominal) inputting the settle down of the coordinates of P & Q, the formula becomes: g(x,h) = (f(x+h)-f(x))/h So, for example, if we input value so that f(x) = x2, and P = (3,9) with h = 0.

2 The formula tells us that g(x,h) at point P = ((x+h)2 x2)÷h becomes P = ((3+0.2)2 32) ÷0.2, which equates to 6.2 The Assignment concealment sheet states that: What you must do is find out what happens to g(x,h) for f(x) = 2 as P takes different! numerical set (P = 1,2,3 ) and as h take downs smaller and smaller [h → 0]. This was done on the ready reckoner: Inputting the formula in calculator terms, and then adding in the ease values of h which tend towards 0, and also setting the values of P starting from 1, then continuing onwards to 2, 3 etc. Inputting values of h tending towards 0...If you want to get a full essay, order it on our website: BestEssayCheap.com

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