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x*x1/2= x3/2 Derivative = 3/2 * x1/2
Suppose you have a function y = f(x) which has an inverse. Therefore there exists a function g(y) such that g(y) = x whenever y = f(x). Now suppose a line parallel to the x axis, y = k (some constant), intersects the graph of y = f(x) at more than one point: say x1 and x2. That means that k = f(x1) and k = f(x2). Now, in the context of the function g, this means that [from the first intersection] g(k) = x1 and [from the first intersection] g(k) = x2 But the function g cannot map k to two different points. That is the contradiction which precludes the possibility of a horizontal line intersecting an invertible function more than once.
44=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.
(x1, y1) = (x - 8, y + 9)
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
G g g f g e e e e x4 abcggfed x1 g g g f g e e e e x4 abcggfed x1 g g g f g e e e e x4 abcggfed x1 g g g f g e e e e x4 abcggfed x1 g g g f g e e e e x4 abcggfed x1 g g g f g e e e e x4 abcggfed x1
it equals x1 it equals x1
Nope.* * * * *The above answer is so wrong!Suppose f and g are two transformations wheref(x) = 2x, andg(x) = x2Then f(g(x)) = f(x2) = 2x2Whileg(f(x)) = g(2x) = (2x)2=4x2Therefore f(g(x)) = g(f(x)) only when x = 0
x*x1/2= x3/2 Derivative = 3/2 * x1/2
x1 = x
Suppose you have a function y = f(x) which has an inverse. Therefore there exists a function g(y) such that g(y) = x whenever y = f(x). Now suppose a line parallel to the x axis, y = k (some constant), intersects the graph of y = f(x) at more than one point: say x1 and x2. That means that k = f(x1) and k = f(x2). Now, in the context of the function g, this means that [from the first intersection] g(k) = x1 and [from the first intersection] g(k) = x2 But the function g cannot map k to two different points. That is the contradiction which precludes the possibility of a horizontal line intersecting an invertible function more than once.
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.
44=2
(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.