x*x1/2= x3/2 Derivative = 3/2 * x1/2
It is x1 which is x.
Here are some examples. x1/2 = square root of x; x1/3 = cubic root of x; in general, x1/n = nth root of x. Also, x2/3 = the square of the cubic root of x, or equivalently, the cubic root of the square of x.
X or if it says what "x" equals then plug that in and multiply
What do you mean by "compute"? Do you want to graph it? Factor it? Calculate it's function given a set of points that lie on it? If you're looking to compute the function given three points that fall on the parabola, then I have just the code for you. If you're given three points, (x1, y1), (x2, y2) and (x3, y3), then you can compute the coefficients of your quadratic equation like this: a = (y1 * (x2 - x3) + y2 * (x3 - x1) + y3 * (x1 - x2)) / (x1 * x1 * (x2 - x3) + x2 * x2 * (x3 - x1) + x3 * x3 * (x1 - x2)) b = (y1 - y2) / (x1 - x2) - a * (x1 + x2); c = y1 - (x1 * x1) * a - x1 * b; You now can calculate the y co-ordinate of any point given it's x co-ordinate by saying: y = a * x * x + b * x + c;
sqr.rtx/x= sqrt.x*sqr.rtx/sqr.rtx=x/x*sqrt.x=1/sqrt.x. x1/2 = x1/2 * x1/2 = x = 1 (x1/2) /x= 1/x1/2
it equals x1 it equals x1
x*x1/2= x3/2 Derivative = 3/2 * x1/2
x1 = x
The proof of the Newton-Raphson iterative equation involves using calculus to show that the method converges to the root of a function when certain conditions are met. By using Taylor series expansion and iterating the equation, it can be shown that the method approaches the root quadratically, making it a fast and efficient algorithm for finding roots.
(y-y1)=m(x-x1) OR we can write it y=m(x-x1)+y1
y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is a point on the line.
It is x1 which is x.
Here are some examples. x1/2 = square root of x; x1/3 = cubic root of x; in general, x1/n = nth root of x. Also, x2/3 = the square of the cubic root of x, or equivalently, the cubic root of the square of x.
X or if it says what "x" equals then plug that in and multiply
2
this is the increasing function theorem, hope it helps "If F'(x) >= 0 , and all x's are and element of [a,b], Then F is increasing on [a,b]" use Mean Value Theorem (M.V.T) Let F'(x)>=0 on some interval Let x1< x2 (points from that interval) by M.V.T there is a point C which is an element of [x1,x2] such that F(x2)-F(x1) / X2- X1 = F'(C) this implies: F(x2)-F(x1) = F'(C) X [x2-x1] F'(C)>=0 [x2-x1]>0 therefore: F(x2)>=F(x1) Therefore: F is increasing on that interval.