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y < 1
It is y >= 5.
The range of the function y=x^2 would be y is greater than or equal to 0 in this case. So pretty much just find the vertex of the function and what ever the y coordinate is set that as the lowest number for the range.
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.
The function y=x is a straight line. The range is all real numbers.The functions just tend to infinity as the x and y values get infinitely large or infinitely small.
Domain (input or 'x' values): -∞ < x < ∞.Range (output or 'y' values): -2 ≤ y ≤ 2.
y = 2sin(x)? If that's your function, well we know that sin(x) oscillates between y = 1 and y = -1, but in our case we have double that from 2sin(x), so our range is -2 to 2.
y < 1
yes y=x Like 2=2
The function y=x is a straight line. The range is all real numbers.
It is y >= 5.
x is a member of the function's domain, y is a member of the function's range.
The range of the function y=x^2 would be y is greater than or equal to 0 in this case. So pretty much just find the vertex of the function and what ever the y coordinate is set that as the lowest number for the range.
The range is the y, while the domain is the x.
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.
The function y=x is a straight line. The range is all real numbers.The functions just tend to infinity as the x and y values get infinitely large or infinitely small.
f ( x ) = (x-2)/(x-1)if y = (x-2)/(x-1)yx-y= x - 2yx-x= -2+yx(y-1)=y-2x = (y-2)/(y-1)so g ( x ) the inverse function is also (x-2)/(x-1)