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

Hint: 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 ${f^{ - 1}}$ .