The domain of the function f (x) = square root of (x - 2) plus 4 is Domain [2, ∞)
Find the domain of the relation then draw the graph.
y=2x+4 --> x=2y+4 ==> y=(x-4)/2
y = 4(2x) is an exponential function. Domain: (-∞, ∞) Range: (0, ∞) Horizontal asymptote: x-axis or y = 0 The graph cuts the y-axis at (0, 4)
If you mean: y-2 = 3x+2 then as a straight line equation it is y = 3x+4
The square root operation is not a function because for each value of y there can be 2 values of x - the principal square root and its negative. This can only be rectified by limiting the range of the opearation to the principal or positive square root. Furthermore, it also depends on the domain of the function. If y<4 then the square root is not defined within Real numbers. So, for y ≥ 4, x = +sqrt(y-4) is a function.
The domain of the function 1/2x is {0, 2, 4}. What is the range of the function?
To find the range of the function ( f(x) = -10x ) for the given domain (-4, -2, 0, 2, 4), we can evaluate the function at each point in the domain. For ( x = -4 ), ( f(-4) = 40 ) For ( x = -2 ), ( f(-2) = 20 ) For ( x = 0 ), ( f(0) = 0 ) For ( x = 2 ), ( f(2) = -20 ) For ( x = 4 ), ( f(4) = -40 ) Thus, the range of the function is ([-40, 40]).
The range is {-7, 1, 9, 17}.
The function ( f(x) = -x^2 + 4 ) is a downward-opening parabola. The vertex, which is the maximum point, occurs at ( x = 0 ) and gives the maximum value of ( f(0) = 4 ). As ( x ) moves away from 0 within the domain, the function decreases, reaching a minimum value at the edges of the domain. If the domain is limited to 201, the range of ( f(x) ) will be from ( -x^2 + 4 ) evaluated at the endpoints of that domain, specifically ( f(201) = -201^2 + 4 = -40400 + 4 = -40396 ). Therefore, the range of ( f(x) ) when the domain is 201 is ( [-40396, 4] ).
The domain is all real numbers except when the denominator equals zero: x2 - 4 = 0 x2 = 4 x = 2, -2 So the domain is all real numbers except 2 and -2.
It is -12.8, -6.4, 0, 6.4 and 12.8
(x + 2)(x + 4) x = -2, -4
A function is a mapping from one set to another. It may be many-to-one or one-to-one. The first of these sets is the domain and the second set is the range. Thus, for each value x in the domain, the function allocates the value f(x) which is a value in the range. For example, if the function is f(x) = x^2 and the domain is the integers in the interval [-2, 2], then the range is the set [0, 1, 4].
x can have any value in the domain.
The mean is synonymous with the average, the sum of the numbers divided by the quantity of the numbers. For example, the average or mean of 2, 4 and 9 is 5, because (2 + 4 + 9) ÷ 3 = 5. I am not familiar with the term domain being a property of a set of constants. I understand a domain to be a property of a mathematical function; specifically, the domain of a function is the set of all possible inputs to the function that yield real individual outputs. For example, if a function of x is 4 ÷ (x - 2), the domain of x is any real number other than 2, since 2 would cause division by zero, so the output would not be a real number.
Yes. Typical example: y = x2. To avoid comparing infinite sets, restrict the function to integers between -3 and +3. Domain = -3, -2 , ... , 2 , 3. So |Domain| = 7 Range = 0, 1, 4, 9 so |Range| = 4 You have a function that is many-to-one. One consequence is that, without redefining its domain, the function cannot have an inverse.
If this is the whole of the function, then the domain is {2, 1, -3, -1}. That set can be put in increasing order if you wish but that is not necessary.