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Question: Find the domain of \(\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}\)...

Find the domain of 1(2−x)2+1\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}

Explanation

Solution

Find points where the given function is not defined
In order to solve this question, the very first step should be to find the points where the function will be defined. Now, we can clearly see that something of the type(2−x)2+1=0\sqrt {{{\left( {2 - x} \right)}^2} + 1} = 0will only be undefined if a=0a = 0. Therefore when we equate (2−x)2+1=0\sqrt {{{\left( {2 - x} \right)}^2} + 1} = 0we get that for no value of x this function will have a value equal to 0. Therefore, we can say that 1(2−x)2+1\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}will never be undefined for any given value of x, hence we can conclude that the domain of the given function is (−∞,∞)\left( { - \infty ,\infty } \right).

Complete step-by-step answer:
The given algebraic expression we have is 1(2−x)2+1\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}
Now, we know that the domain of a function is the range of possible values it can take. If we put all the possible numbers in it
Let’s say our function is f(x)=2xf(x) = 2x
Now, for all the possible real numbers x, we will get infinitely different values of f(x)f(x)
This clearly means that, f(x)=2xf(x) = 2xcan take almost every possible value for every possible value of x
Hence, we will say that, f(x)=2xf(x) = 2xdomain is (−∞,∞)\left( { - \infty ,\infty } \right)
Now, let’s take f(x)=1x−1f(x) = \dfrac{1}{{x - 1}}
Clearly, f(x)f(x)doesn’t have any value at x=1. Because then it will become 1∞\dfrac{1}{\infty }which is not defined
except that, f(x)f(x)has a value for every other number
Hence we will say that, domain of f(x)=1x−1f(x) = \dfrac{1}{{x - 1}}is (−∞,1)∪(1,∞)\left( { - \infty ,1} \right) \cup \left( {1,\infty } \right)
Therefore, if we now solve this problem, we can clearly say that for f(x)=1(2−x)2+1f(x) = \dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}
f(x)f(x)is defined for every possible real number as (2−x)2+1\sqrt {{{\left( {2 - x} \right)}^2} + 1} can never become 0
Hence, for every possible value of x, there is a value off(x)f(x).
So, we will conclude that, domain of f(x)=1(2−x)2+1f(x) = \dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}is(−∞,∞)\left( { - \infty ,\infty } \right).

Note: Students often get confused between domain and codomain. Domain is the set of possible values, the functions can take an input for. Whereas, codomain is the set of values which the function can give for any possible value of x as input. Basically domain is the possible values of x, and codomain is the possible values of y.