1-2 is not a function because there is no defined function, such as f(n)=0n+1-2. Also, it does not contain a variable.
Yes, a function can have a negative 1 output if there is an input value for which the function evaluates to -1. For example, the function f(x) = x - 2 has an output of -1 when the input x is 1, since f(1) = 1 - 2 = -1. The specific outputs of a function depend on its definition and the values in its domain.
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.
To find the vertex of the quadratic function ( y = x^2 - 4x + 1 ), we can use the vertex formula ( x = -\frac{b}{2a} ), where ( a = 1 ) and ( b = -4 ). This gives ( x = -\frac{-4}{2 \cdot 1} = 2 ). Substituting ( x = 2 ) back into the equation, we find ( y = 2^2 - 4(2) + 1 = -1 ). Thus, the vertex of the function is at the point ( (2, -1) ).
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! + . . . )
Any multiple of X^2+X/2-1/2
The function t(n) relates to the function t(n1/2) 1 by taking the square root of n in the second function and adding 1 to the result.
Yes, a function can have a negative 1 output if there is an input value for which the function evaluates to -1. For example, the function f(x) = x - 2 has an output of -1 when the input x is 1, since f(1) = 1 - 2 = -1. The specific outputs of a function depend on its definition and the values in its domain.
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.
To find the vertex of the quadratic function ( y = x^2 - 4x + 1 ), we can use the vertex formula ( x = -\frac{b}{2a} ), where ( a = 1 ) and ( b = -4 ). This gives ( x = -\frac{-4}{2 \cdot 1} = 2 ). Substituting ( x = 2 ) back into the equation, we find ( y = 2^2 - 4(2) + 1 = -1 ). Thus, the vertex of the function is at the point ( (2, -1) ).
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).