I read all comments, thanks for your quick replies!
@krb17:
for my example:
y = x^2 ; given x = +(-)2 => y (looking for) = 4
x = +(-)sqrt(y) ; given y = 4 => x (looking for) = +(-) 2
The purpose of the inverse function was that we want to find the value of x when the value of y was...
@hunt mat: can you explain that to me with an example ?
@JJacquelin:
I made a graph of what you wrote:
y=x^2 is the explicit function (purple)
x-sqrt(y) = 0 is the implicit function (red)
those are the same functions...
I think the inverse function is: f-1(y)=y^2 =x
because...
Hello,
I'am suffering with the theoretical background. My course state the follow thing:
D(f-1(y))=1/D(f(x)).
So: f-1(y) is the inverse function of f(x), this means that the argument of f-1(y) is y!
Example: y = f(x) = x^2 => f-1(y): x = y^2. Am I correct with is one ?
The chain...
Oké, but there exist an equation in polar form to describe the cartesian equation ? Because there al multiple x,y coordinates, and one of them wil point to the cartesian equation as you explained...
greetz
Hello,
thank you all for quick replies !
The cartesian equation y = 2.sin(x)-3 has indeed x as argument; ofcourse in radians.
But what I know form transformation to polar form is that the y x plane transforms to a
r \theta plane. Being \theta tha angle between r and de x axle. In my case...
Hello,
I'am new here and happy to find this great forum!
Here's my first question: there's an explicit function as follows:
y=2.sin(x)-1
The transformation to polar form (r=3cos(\theta))
- x=r.cos(\theta)
- y=r.sin(\theta)
So I get: r.sin(\theta)=2.sin(r.cos(\theta))-1
Now you see...