f(71.19) = 71 where int(x) is the integer value of a number
It i the set of values that f(x) can take for those value of x. However, since you have not bothered to share information on what f(x) is, I cannot provide a more useful answer.
It depends on whether fx denotes frequency times variable value or the probability generating function for the variable x.
That depends what the function is! Just replace every occurrence of "x" by "x-2", and do any required simplifications.
If x and f are quantities then xf and fx are the same if they belong to a commutative space. If not, they normally do not. Multiplication of matrices, for example is cont commutative. In fact, it is quite possible that xf has a value but fx does not even exist. if x and f are functions then again, fx and xf will not generally be the same. For example suppose x = "add 5 to it" and f = "square it". Then xf(3) = x(9) = 14 while fx(3) = f(8) = 64
You replace x with 1, and 36 times 1 = 36.
f(71.19) = 71 where int(x) is the integer value of a number
It i the set of values that f(x) can take for those value of x. However, since you have not bothered to share information on what f(x) is, I cannot provide a more useful answer.
It depends on whether fx denotes frequency times variable value or the probability generating function for the variable x.
That depends what the function is! Just replace every occurrence of "x" by "x-2", and do any required simplifications.
At x = 3, the value of F(x) = 3x + 2 is the value 11, which graphs to the point (3, 11).
If x and f are quantities then xf and fx are the same if they belong to a commutative space. If not, they normally do not. Multiplication of matrices, for example is cont commutative. In fact, it is quite possible that xf has a value but fx does not even exist. if x and f are functions then again, fx and xf will not generally be the same. For example suppose x = "add 5 to it" and f = "square it". Then xf(3) = x(9) = 14 while fx(3) = f(8) = 64
f(x) = x2 + 5x + 6At a critical point, f'(x) = 0.f'(x) = 2x + 5 = 02x = -5x = -2.5f(-2.5) = (-2.5)2 + 5(-2.5) + 6 = (6.25) - (7.5) + (6) = 4.75
y = f(x) = (x-5)2. At x = 5, the y value is zero, so the point (5,0).
Negative the derivative of f(x), divided by f(x) squared. -f'(x) / f²(x)
[fx] is a function of x, it usually used in graphs.
If you mean: 6 = x-1 then the value of x is 7