Question

If $\mathbb{R}$ be the set of all real numbers and $f:\mathbb{R} \to \mathbb{R}$ is given by $f\left( x \right) = 3{x^2} + 1$ . Then, the set of ${f^{ - 1}}\left( {\left[ {1,6} \right]} \right)$ is
A) \left\\{ { - \sqrt {\dfrac{5}{3}} ,0,\sqrt {\dfrac{5}{3}} } \right\\}
B) $\left[ { - \sqrt {\dfrac{5}{3}} ,\sqrt {\dfrac{5}{3}} } \right]$
C) $\left[ { - \sqrt {\dfrac{1}{3}} ,\sqrt {\dfrac{1}{3}} } \right]$
D) $\left( { - \sqrt {\dfrac{5}{3}} ,\sqrt {\dfrac{5}{3}} } \right)$

Answer 1

The inverse of the function ${f^{ - 1}}$ exists only if the function $f\left( x \right)$ is one-one and onto.
One-one function: When each element of set B is mapped to only one element of set A, i.e., each object in set A has a unique image in set B, then the function is called one-one function.
onto functions. If for functions $f:{\text{A}} \to {\text{B}}$ the co-domain set of B is also the range for the function, then the function is called an onto function.

Complete step-by-step answer:
Step 1: Check if the given function is one-one.
Given that for $f:\mathbb{R} \to \mathbb{R}$ the function $f\left( x \right) = 3{x^2} + 1$
Where $\mathbb{R}$ is the set of all real numbers.
Let ${x_1},{x_2} \in \mathbb{R}$ (that is the domain of the given function).
The function is one-one for $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$

$$\Rightarrow 3{x_1}^2 + 1 = 3{x_2}^2 + 1 \\\ \Rightarrow 3{x_1}^2 = 3{x_2}^2 \\\ \Rightarrow {x_1}^2 = {x_2}^2 \\\$$

Since for ${x_1} = 1 \in \mathbb{R}$ ; ${x_1}^2 = 1$
And for ${x_2} = - 1 \in \mathbb{R}$ ; ${x_2}^2 = 1$
We have, ${x_1}^2 = {x_2}^2$ but ${x_1} \ne {x_2}$
Thus the given function $f\left( x \right) = 3{x^2} + 1$ is not one-one, hence the function is not invertible. Thus, the inverse of the function does not exist for the co-domain $\mathbb{R}$ of the function $f\left( x \right)$. So none of the options is correct.
But if we redefine the domain for $x \geqslant 0$ the function will become the one-one. But still, the function is not onto so we will have to redefine the co-domain as well.
The output values of the function $f\left( x \right) = 3{x^2} + 1$ will always be positive as the square of the x is involved.
For x = 0, $f\left( 0 \right) = 3\left( {{0^2}} \right) + 1 = 0 + 1 = 1$
For x = 1, $f\left( 1 \right) = 3\left( {{1^2}} \right) + 1 = 3 + 1 = 4$
For x = -1, $f\left( { - 1} \right) = 3\left( { - {1^2}} \right) + 1 = 3 + 1 = 4$
Thus, the range of the function $f\left( x \right) = 3{x^2} + 1$ is $R = \left[ {1,\infty } \right)$
The graphical representation of the $f\left( x \right) = 3{x^2} + 1$with its domain $x \in \left[ {0,\infty } \right)$ and its co-domain $f\left( x \right) \in \left[ {1,\infty } \right)$

So the new function will be $f:\left[ {0,\infty } \right) \to \left[ {1,\infty } \right),f\left( x \right) = 3{x^2} + 1$.
Now the function is invertible.
$f\left( x \right) = 3{x^2} + 1$
Differentiating both sides with respect to x.
$f'\left( x \right) = 6x$
$f'\left( x \right)$ is positive, hence the function $f\left( x \right)$ is an increasing function. Therefore we will just have to calculate the values at the endpoint only in ${f^{ - 1}}\left( x \right)$
Step 2: let's find the inverse of the given function in the redefined domain.
$y = 3{x^2} + 1 \\\ \Rightarrow 3{x^2} = y - 1 \\\ \Rightarrow {x^2} = \dfrac{{y - 1}}{3} \\\ \Rightarrow x = \pm \sqrt {\dfrac{{y - 1}}{3}} \\\$
Interchange $x \leftrightarrow y$
$\Rightarrow y = \pm \sqrt {\dfrac{{x - 1}}{3}}$
Thus ${f^{ - 1}}\left( x \right) = \pm \sqrt {\dfrac{{x - 1}}{3}}$
Step 3: Find the set of ${f^{ - 1}}\left( {\left[ {1,6} \right]} \right)$
${f^{ - 1}}\left( x \right) = \sqrt {\dfrac{{x - 1}}{3}}$
For x = 1, ${f^{ - 1}}\left( 1 \right) = \sqrt {\dfrac{{1 - 1}}{3}} = 0$
For x = 6, ${f^{ - 1}}\left( 6 \right) = \sqrt {\dfrac{{6 - 1}}{3}} = \sqrt {\dfrac{5}{3}}$
${f^{ - 1}}\left( {\left[ {1,6} \right]} \right) \in \left[ {0,\sqrt {\dfrac{5}{3}} } \right]$

Similarly, if we redefine the domain as $x \in \left( { - \infty ,0} \right]$ and codomain as the $\left[ {1,\infty } \right)$ we will get ${f^{ - 1}}\left( {\left[ {1,6} \right]} \right)$
${f^{ - 1}}\left( x \right) = - \sqrt {\dfrac{{x - 1}}{3}}$
For x = 1, ${f^{ - 1}}\left( 1 \right) = - \sqrt {\dfrac{{1 - 1}}{3}} = 0$
For x = 6, ${f^{ - 1}}\left( 6 \right) = - \sqrt {\dfrac{{6 - 1}}{3}} = - \sqrt {\dfrac{5}{3}}$
we get ${f^{ - 1}}\left( {\left[ {1,6} \right]} \right) \in \left[ { - \sqrt {\dfrac{5}{3}} ,0} \right]$
Thus, ${f^{ - 1}}\left( {\left[ {1,6} \right]} \right)$ $\in \left[ { - \sqrt {\dfrac{5}{3}} ,\sqrt {\dfrac{5}{3}} } \right]$ for different domains.
Final answer: The set of ${f^{ - 1}}\left( {\left[ {1,6} \right]} \right)$ is $\left[ { - \sqrt {\dfrac{5}{3}} ,\sqrt {\dfrac{5}{3}} } \right]$.

Thus, the correct option maybe (C).

Note: The composition of functions is also useful in checking the invertible function, hence finding the inverse of the function.
A function $f:X \to Y$ is defined to be invertible, if there exists a function $g:Y \to X$ such that $gof = {I_x}$and $fog = {I_y}$ . The function $g$ is called the inverse of $f$ and is denoted by ${f^{ - 1}}$ .