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.
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)
Ans: A natural log function ALWAYS has base e ( e is the irrational number that is the sum of the infinite series 2 + 1 / 2! + 1 /3! + 1 /4! + . . . )
y = 1/2 x + 1
{(5, 2), (3, 4), (1, 2), (−1, 4)}
Any multiple of X^2+X/2-1/2
2
1) Function 2) Which will excute particular methof/function
1) Function 2) Which will excute particular methof/function
Y = x2
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.
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)
Ans: A natural log function ALWAYS has base e ( e is the irrational number that is the sum of the infinite series 2 + 1 / 2! + 1 /3! + 1 /4! + . . . )
The function sec(x) is the secant function. It is related to the other functions by the expression 1/cos(x). It is not the inverse cosine or arccosine, it is one over the cosine function. Ex. cos(pi/4)= sqrt(2)/2 therefore secant is sec(pi/4)= 1/sqrt(2)/2 or 2/sqrt(2).
The domain of the function 1/2x is {0, 2, 4}. What is the range of the function?
The answer will range between '2' & '-2' Reason; The Sine function ranges between '1' & '-1' , so if it has a coefficient of '2', this will increase the range to '2' & '-2'.
A function of x means that the answers depends on the value of x that is substituted into the function. For example: y = x + 1 If x = 1, y = 2 If x = 2, y = 3 and so on.