Q: How do you find the domain and range of f ( x ) x 2 1?

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sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].

The range of cosine is [-1, 1] which is, therefore, the domain of cos-1. As a result, cos-1(2) is not defined.

The domain would be (...-2,-1,0,1,2...); the range: (12)

y=x^2

The simplest answer is that the domain is all non-negative real numbers and the range is the same. However, it is possible to define the domain as all real numbers and the range as the complex numbers. Or both of them as the set of complex numbers. Or the domain as perfect squares and the range as non-negative perfect cubes. Or domain = {4, pi} and range = {8, pi3/2} Essentially, you can define the domain as you like and the definition of the range will follow or, conversely, define the range and the domain definition will follow,

Related questions

The domain of the function 1/2x is {0, 2, 4}. What is the range of the function?

The answer depends on the domain. If the domain is the whole of the real numbers, the range in y â‰¥ 1. However, you can choose to have the domain as [1, 2] in which case the range will be [2, 5]. If you choose another domain you will get another range.

sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].

the domain is all real numbers and the range is all real numbers the domain is all real numbers and the range is all real numbers

The range could be anything. Without parameters specified, the domain of {1,2,3,4} could have any range. This problem is unsolvable.

That depends on the specific function.

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.

The range is {-5, -2, 1, 4}

The range of cosine is [-1, 1] which is, therefore, the domain of cos-1. As a result, cos-1(2) is not defined.

x y -3 2 -1 6 1 -2 3 5

The domain would be (...-2,-1,0,1,2...); the range: (12)

Given: f(x)=3x^2+6x-2 To find x: x= -b/2a x= -6/2(3) x= -1 to find y, replace x with -1: f(-1)=3(-1)^2+6(-1)-2 y=3(1)-6-2 y= -5