Question

If \[f:\left[ {1,\infty } \right) \to \left[ {2,\infty } \right)\] is given by \[f\left( x \right) = x + \dfrac{1}{x}\] then find \[{f^{ - 1}}\left( x \right)\].
A. \[\dfrac{{x + \sqrt {{x^2} - 4} }}{2}\]
B. \[\dfrac{x}{{1 + {x^2}}}\]
C. \[\dfrac{{x - \sqrt {{x^2} - 4} }}{2}\]
D. \[1 + \sqrt {{x^2} - 4} \]

Answer 1

Hint: The inverse function will return the initial value wherein the output of a function was given. In this question, need to find the inverse of a function \[f\left( x \right)\] . . After simplification, based on the given range, we can find \[{f^{ - 1}}\left( x \right)\]. ]

Complete step-by-step answer: Let \[y = f\left( x \right)\]
Thus, we get
\[
  y = x + \dfrac{1}{x} \\
   \Rightarrow xy = {x^2} + 1 \\
   \Rightarrow {x^2} - xy + 1 = 0 \\
 \]
Let us factorize the above quadratic equation \[{x^2} - xy + 1 = 0\]
It is in the form of \[a{x^2} + bx + c = 0\]
Thus, by comparing the above equation with \[{x^2} - xy + 1 = 0\], we get \[a = 1;b = - y;c = 1\]
So, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
By putting the values of \[a = 1;b = - y;c = 1\] in the above equation, we get
Hence, \[x = \dfrac{{ - \left( { - y} \right) \pm \sqrt {{{\left( { - y} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}\]
\[x = \dfrac{{y \pm \sqrt {{{\left( y \right)}^2} - 4} }}{2}\]
Now, we can have \[{f^{ - 1}}\left( y \right) = \dfrac{{y \pm \sqrt {{{\left( y \right)}^2} - 4} }}{2}\]
\[{f^{ - 1}}\left( x \right) = \dfrac{{x \pm \sqrt {{{\left( x \right)}^2} - 4} }}{2}\]
We know that the range of a inverse function is \[\left[ {1,\infty } \right)\]
If we take \[{f^{ - 1}}\left( x \right) = \dfrac{{x - \sqrt {{{\left( x \right)}^2} - 4} }}{2}\] then \[{f^{ - 1}}\left( x \right) > 1\]
So, this is possible if and only if
\[
  {\left( {x - 2} \right)^2} > {x^2} - 4 \\
   \Rightarrow {x^2} - 2\left( 2 \right)\left( x \right) + {\left( 2 \right)^2} + > {x^2} - 4 \\
   \Rightarrow {x^2} - 4x + 4 + > {x^2} - 4 \\
   \Rightarrow - 4x + 8 > 0 \\
 \]
By simplifying further, we get
\[
   \Rightarrow 8 < 4x \\
   \Rightarrow 2 < x \\
 \]
Where \[x > 2\]
So, consider \[{f^{ - 1}}\left( x \right) = \dfrac{{x + \sqrt {{{\left( x \right)}^2} - 4} }}{2}\]


Therefore, the correct option is (A).

Note: Many students make mistakes in the calculation part. This is the only way through which we can solve this example in an easy manner.