No, but cos(-x) = cos(x), because the cosine function is an even function.
y < 1
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
3
y = x2 + x = 0 x (X + 1) = 0 x = 0 is one solution x = -1 is the other
f(f(x)) = f(x). Only if f is 1-1 then we have a solution f(x)=x.
No, but cos(-x) = cos(x), because the cosine function is an even function.
f(x) = 2*(x-3)*(x+2)/(x-1) for x ≠1
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Yes. Think of y as being a function of x. y = f(x) = x2 + 1
y < 1
X - Y^2 = 1 - Y^2 = - X + 1 Y^2 = X - 1 Y = (+/-) sqrt(X - 1) now, X is represented as a function of Y. Function values are generally Y values.
3
y = x2 + x = 0 x (X + 1) = 0 x = 0 is one solution x = -1 is the other
2x-2/x^2+3x-4
It is a linear function. That is to say, it is a function representing a straight line in the coordinate plane.
What is the area bounded by the graph of the function f(x)=1-e^-x over the interval [-1, 2]?