Hint: We can find the domain by finding the value of x that makes the function undefined and subtracting it from the set of real numbers. For that we equate the denominator to 0 and solve for x. For finding the range, we can write the interval of . Then we can transform it to the function by addition, multiplication and taking reciprocals, to get the range of the function.Complete step-by-step answer:We know that domain of a function is the set of all the possible values that the variable in that function takes. We know that a real valued function is not defined when the denominator becomes zero and the term inside the root is negative.Here, we have the function . The function has a denominator. So, the range must not have the value which makes the denominator 0. We can equate the denominator to 0 and solve it to get that value of x.  On rearranging, we get,  On taking the square root, we get,  So, x cannot take the values and .Therefore, the domain of the function is set of all the real numbers except the numbers and . $ \\Rightarrow D = R - \\left\\{ {\\; - \\sqrt 2 ,\\sqrt 2 } \\right\\}$ Range of a function is the set of all the values that the function gives for the values of variables in the domain.  We know that, On multiplying with -1, the equality reverses,\\[ \\Rightarrow \\] On adding 2, we get,\\[ \\Rightarrow \\] According to the domain of the function, . So cannot take value 0.   and On taking the reciprocal, we get,  On multiply throughout with 3, we get,  We know that and   So, y can take values in the interval Therefore, the domain of the function is $R - \\left\\{ {\\; - \\sqrt 2 ,\\sqrt 2 } \\right\\}$ and range is .Note: Domain of a function is the set of all the values that the variable can take. Range of a function is the set of all the values that the function gives for the values of variables in the domain. While calculating the range we must consider that in the range of the function and change the interval accordingly. To verify our answer, we can check what are the values of y when tends to infinity, 2, zero and negative infinity.

Hint: According to Set theory a function is defined as a relation between two sets A and B such that every element of the first set has exactly one image in the second set.Complete step by step solution: If any element of the first set does not have an image in the second set, or if an element in the first set has more than one image in the second set, then the relation Is not a function from A to B.The first set which contains the elements $'x'$ is called the Domain of the function $f$ while the second set which contains the images defined as per the function $f$, that is,$f(x)$is called the range of the function $f$.For example:\n \n \n \n \n Now, let us take two sets A and B respectively containing the elements A =$\\left\\{ {C,D,E,F,G} \\right\\}$and B = $\\left\\{ {2,3,5,7,9} \\right\\}$such that for all elements $'x'$ of the domain A there exists the image elements $f(x)$ in B under the function $f$, respectively. Such that:$ f\\left( C \\right) = 2 \\\\  f\\left( D \\right) = 5 \\\\  f\\left( E \\right) = 3 \\\\  f\\left( F \\right) = 9 \\\\  f\\left( G \\right) = 7 \\\\ $However the function $f$, will not be a function if any element of the Domain is associated with more than one element in the range. For example,\n \n \n \n \n Now, the function $f$ is not a function because the element E $ \\in $Domain set A, has two images at 3 and 7 in the Range B. Hence $f$ is not a function.Therefore, we conclude by defining a function $f$as a rule from one set A to another set B, such that the first set A containing the elements $'x'$, is called the Domain , while the set B containing the corresponding images $f(x)$is called the Range of the function $f$.Note: Functions are of many types. Let us consider two sets A and B. A one-to-one function is one where every element belonging to a set A, that has exactly one unique image in set B. An onto function is one where every element belonging to set B has at least one preimage in the set A. A function which is both one-to-one and onto, is called an injective function. If either of the characteristics is missing, the function will not be an injective function.