x<=8 is the domain of G(F(x))
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That is true only if the range of G is the set of real numbers. If the range of G is C,the complex field (and why should it not be?) than the domain of G can be any subset of C.
Assuming the standard x and y axes, the range is the maximum value of y minus minimum value of y; and the domain is the maximum value of x minus minimum value of x.
The domain of the function f(x) = (x + 2)^-1 is whatever you choose it to be, except that the point x = -2 must be excluded. If the domain comes up to, or straddles the point x = -2 then that is the equation of the vertical asymptote. However, if you choose to define the domain as x > 0 (in R), then there is no vertical asymptote.
The square root of 400 is 20, as 20 multiplied by 20 equals 400. The square root of 196 is 14, as 14 multiplied by 14 equals 196. Therefore, the expression simplifies to 20 - 14, which equals 6.
So u have the function x2-4=0 Now because 4 is a squared number, its square root is 2 (x+2)(x-2)=0
D = {x [element of reals]}R = {y [element of reals]|y >= 4}
The domain could be the real numbers, in which case, the range would be the non-negative real numbers.
33/4
The domain is what you choose it to be. You could, for example, choose the domain to be [3, 6.5] If the domain is the real numbers, the range is [-12.25, ∞).
y=(√1)-x2 The domain is the set of numbers that "x" can be. In this equation "x" can be any real number. The domain for this problem would be (-inf,inf) *Inf= Infinity*
The domain and the range depends on the context. For example, the domain and the range can be the whole of the complex field. Or I could define the domain as {-2, 1, 5} and then the range would be {0, 3, -21}. When either one of the range and domain is defined, the other is implied.
it is greater than and equal to 3
Plus or minus the base. If the base is X and you square it, you get X2. If you take the square root of that, you get Plus or Minus X. This is because X*X equals X2 and -X*-X also equals X2.
No. But sin2a equals 1 minus cos2a ... and ... cos2a equals 1 minus sin2a
The are under the curve on the domain (a,b) is equal to the integral of the function at b minus the integral of the function at a
Only if the domain (the numbers that you put into the function) are "bigger than or equal to zero". If the domain is "all real numbers",(i.e. including negatives) then it is not a one-to-one function. The question will tell you what the domain of the function (i.e. the values of 'x' that you are meant to input).
If x equals the square root of ...., then you already have solved for x
square root -5 minus 14 or - square root -5 minus 14