Question

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

Answer 1

Hint: Use the basic rule of logarithmic, which is to change the base of logarithmic and remove the above of the expression. Also, the domain of the function is only positive value and you could ignore the negative value.

Complete step-by-step solution:
From the question we know that the function is $f\left( x \right) = {2^{x\left( {x - 1} \right)}}$ and its range is \[f:\left[ {1,\;\infty } \right] \to \left[ {1,\;\infty } \right)\].
Now we let the function $f\left( x \right)$ as $y$ and simplify the equation.
$
\Rightarrow y = {2^{x\left( {x - 1} \right)}}\\
\Rightarrow y = {2^{{x^2} - x}}
$
Now we take logarithm on both sides of the above equation, we get,
$
\Rightarrow \ln \left( y \right) = \ln \left( {{2^{{x^2} - x}}} \right)\\
\Rightarrow \ln y = \left( {{x^2} - x} \right)\ln 2
$
Now we change the base of logarithmic to make calculations easy.
${\log _2}y = {x^2} - x$
Now rewrite the above equation make complete square on right hand side, we get,
$
\Rightarrow {\log _2}y = {\left( {x - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4}\\
\Rightarrow {\log _2}y + \dfrac{1}{4} = {\left( {x - \dfrac{1}{2}} \right)^2}
$
Now we take square root on both side and simplify it, we get,
$ \pm \dfrac{1}{2}\sqrt {1 + 4{{\log }_2}y} = x - \dfrac{1}{2}$
We know from the given range \[f:\left[ {1,\;\infty } \right] \to \left[ {1,\;\infty } \right)\] that the domain is $\left[ {1,\;\infty } \right)$. Therefore, the domain is the only positive and negative sign that can be neglected from the above equation.
$
\Rightarrow \dfrac{1}{2}\sqrt {1 + 4{{\log }_2}y} = x - \dfrac{1}{2}\\
\Rightarrow x = \dfrac{1}{2}\left\{ {1 + \sqrt {1 + 4{{\log }_2}y} } \right\}
$
Now we replace $x$ and $y$, we get,
\[{f^{ - 1}}\left( x \right) = x = \dfrac{1}{2}\left\{ {1 + \sqrt {1 + 4{{\log }_2}x} } \right\}\]

Hence, the inverse of the given function is \[{f^{ - 1}}\left( x \right) = x = \dfrac{1}{2}\left\{ {1 + \sqrt {1 + 4{{\log }_2}x} } \right\}\].

Additional Information: The domain of a function is all the values by which the given function is satisfied and domain is the expansion on x axis whereas the range is the all the values that the given function takes and range is the expansion on y axis.

Note: The formula used for changing the base of the logarithmic is ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$.
The inverse function means is a function such that for every value of $x$ that $f$ leads to $f\left( x \right)$, but ${f^{ - 1}}$ leads to $f\left( x \right)$ and back to $x$. In inverse relation, the range becomes domain and the domain becomes range.