Question

Check the following whether a function from $\mathbb{R}$ to $\mathbb{R}$ or NOT
a) $f\left( x \right) = \sqrt x $
b) $f\left( x \right) = \sqrt[3]{x}$
c) $f\left( x \right) = \sqrt {x - 2} + \dfrac{1}{{\sqrt {2 - x} }}$
d) ${f^2}\left( x \right) = x$

Answer 1

Hint: We are to individually check every function given above to see whether the domain of the function is $\mathbb{R}$ or NOT, as in many functions a particular domain has to be determined as for the rest of the numbers the function will be undefined. If the domain is $\mathbb{R}$, then we have to see if the range is also $\mathbb{R}$ or not. If so, the function that satisfies both the conditions will be a function from $\mathbb{R}$ to $\mathbb{R}$, or else NOT.

Complete step by step answer:
(a) $f\left( x \right) = \sqrt x $
In this function, $x > 0$, as inside the root the number can’t be negative, i.e., $ < 0$.
Therefore, $x \in {\mathbb{R}^ + }$
That is, the domain is not $\mathbb{R}$.
Therefore, $f\left( x \right) = \sqrt x $ is NOT $\mathbb{R}$ to $\mathbb{R}$.

(b) $f\left( x \right) = \sqrt[3]{x}$
In this function, the domain is $\mathbb{R}$, as we can put any number in $\sqrt[3]{{}}$, and it will provide us with a number.
Also, the range of the function will also be $\mathbb{R}$, as the cube root function will return any value.
Hence, $f\left( x \right) = \sqrt[3]{x}$ is a $\mathbb{R}$ to $\mathbb{R}$ function.

(c) $f\left( x \right) = \sqrt {x - 2} + \dfrac{1}{{\sqrt {2 - x} }}$
We know, inside the root the numbers can’t be negative, i.e., $ < 0$.
So, $x - 2 > 0$
$ \Rightarrow x > 2$
Also, $2 - x > 0$
$ \Rightarrow 2 > x$
Here, both of the inequalities contradict each other.
There is no term in the domain of $x$.
That is, $x \in \phi $ (null set).
Therefore, the domain of $x$ is not $\mathbb{R}$.
Hence, the function $f\left( x \right) = \sqrt {x - 2} + \dfrac{1}{{\sqrt {2 - x} }}$ is NOT $\mathbb{R}$ to $\mathbb{R}$.

(d) ${f^2}\left( x \right) = x$
Now, taking square root on both sides, we get,
$ \Rightarrow f\left( x \right) = \sqrt x $
This is the same as the function (a).
So, in option (a), we already saw that this function is NOT $\mathbb{R}$ to $\mathbb{R}$.
Therefore, this function ${f^2}\left( x \right) = x$ is also NOT $\mathbb{R}$ to $\mathbb{R}$.

Note:
> Every function is not defined in every domain, for some particular values the function may be undefined or we can say we simply can put every value in the function as domain as they will show up with results like $\dfrac{0}{0},\dfrac{1}{0},\infty $ etc. So, we must check whether we are able to put a particular value in the function as a domain.
> A cube root function's domain is the set of all real numbers. Because three negatives equal a negative, a cube root function can employ all real numbers, unlike a square root function, which is confined to nonnegative values.