Question

If the function $f:\left[ {1,\infty } \right) \to \left[ {1,\infty } \right)$ is defined by $f\left( x \right) = {2^{x\left( {x - 1} \right)}}\;\;$is invertible, then ${f^{ - 1}}\left( x \right)\;$ is

Answer 1

Here first we will let the given function to be y and then take log of both the sides and find the value of x in terms of y to get the value of ${f^{ - 1}}\left( y \right)\;$. Then we will replace y by x to get the desired answer.

Complete step-by-step answer:
Let $f\left( x \right) = y$
Then,
$\;x = {f^{ - 1}}\left( y \right)$………………………………..(1)
Also,
$y = {2^{x\left( {x - 1} \right)}}\;$
Now taking log of the sides using the formula:-
$\log \left( {{a^b}} \right) = b\log a$
We get:-
$\log y = x\left( {x - 1} \right).\log 2$
Now we will solve the above equation for the value of x:

$$\log y = \left( {{x^2} - x} \right)\log 2 \\\ \log y = {x^2}\log 2 - x\log 2 \\\ {x^2}\log 2 - x\log 2 - \log y = 0.......................\left( 2 \right) \\\$$

Now, since a quadratic equation is formed , therefore we will use the quadratic formula to find the value of x.
For any quadratic equation of the form $a{x^2} + bx + c = 0$ the quadratic formula is given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing equation 2 with the standard equation we get:-

$$a = \log 2 \\\ b = - \log 2 \\\ c = - \log y \\\$$

Hence applying quadratic formula for these values we get:-

$$x = \dfrac{{ - \left( { - \log 2} \right) \pm \sqrt {{{\left( { - \log 2} \right)}^2} - 4\left( {\log 2} \right)\left( { - \log y} \right)} }}{{2\left( {\log 2} \right)}} \\\ x = \dfrac{{\log 2 \pm \sqrt {{{\log }^2}2 + 4\log 2.\log y} }}{{2\log 2}} \\\$$

Now taking $\log 2$ common from both the numerator and the denominator we get:-

$$x = \dfrac{{\log 2\left[ {1 \pm \sqrt {1 + 4\left( {\dfrac{{\log y}}{{\log 2}}} \right)} } \right]}}{{2\log 2}} \\\ x = \dfrac{{\left[ {1 \pm \sqrt {1 + 4\left( {\dfrac{{\log y}}{{\log 2}}} \right)} } \right]}}{2} \\\$$

Now we will use the following property of log:-
$\dfrac{{\log ea}}{{\log eb}} = \dfrac{{\log ba}}{{\log bb}}$
Now since, $\log bb = 1$
Therefore,
$\dfrac{{\log ea}}{{\log eb}} = \log ba$
Hence applying this property we get:-
$x = \dfrac{{1 \pm \sqrt {1 + 4\log 2y} }}{2}$
Now since, $x \in \left[ {1,\infty } \right)$ therefore, ‘-ve’ sign can be omitted
Hence we get:-
$x = \dfrac{{1 + \sqrt {1 + 4\log 2y} }}{2}$
Now putting this value in equation (1) we get:-
${f^{ - 1}}\left( y \right) = \dfrac{{1 + \sqrt {1 + 4\log 2y} }}{2}$
Replacing y by x we get:-
${f^{ - 1}}\left( x \right) = \dfrac{{1 + \sqrt {1 + 4\log 2x} }}{2}$

Note: The quantity inside the logarithm function can never be zero as logarithm function is not defined at zero
Also, the logarithm function is a strictly increasing function.
Any function f is invertible if and only if it is one – one and onto and if it is invertible then only we can find its inverse