Question

The inverse of \[f(x) = \dfrac{2}{3}\dfrac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}\] is
A. \[\dfrac{1}{3}{\log _{10}}\dfrac{{1 + x}}{{1 - x}}\]
B. \[\dfrac{1}{2}{\log _{10}}\dfrac{{2 + 3x}}{{2 - 3x}}\]
C. \[\dfrac{1}{3}{\log _{10}}\dfrac{{2 + 3x}}{{2 - 3x}}\]
D. \[\dfrac{1}{6}{\log _{10}}\dfrac{{2 - 3x}}{{2 + 3x}}\]

Answer 1

Hint: To solve for the inverse of the second function, we need to convert one function into another using inverse functions. It is necessary to deal with issues involving variables or systems of equations. Linear regression models, which are utilized in forecasting or prediction, can also be solved using inverse functions.
 The function g is the inverse of function \[f\] if for\[y = f\left( x \right)\],\[x = g\left( y \right)\].

Complete step by step solution: A function's domain and range must be determined before you can calculate its inverse. The set of all x-values for which the function returns a value is known as the domain. The collection of all y-values for which the function yields a correct result is known as the range.
By switching \[x\] and \[y\], you can obtain the inverse function by resolving the equation that results.
If the beginning function is not one-to-one, then there will be more than one inverse.
So, swap the variables:
The expression \[y = \dfrac{{\dfrac{{2 \cdot {{10}^x}}}{3} - \dfrac{{2 \cdot {{10}^{ - x}}}}{3}}}{{{{10}^x} + {{10}^{ - x}}}}\] becomes \[x = \dfrac{{\dfrac{{2 \cdot {{10}^y}}}{3} - \dfrac{{2 \cdot {{10}^{ - y}}}}{3}}}{{{{10}^y} + {{10}^{ - y}}}}\].
Solve the equation:
The expression \[x = \dfrac{{\dfrac{{2 \cdot {{10}^y}}}{3} - \dfrac{{2 \cdot {{10}^{ - y}}}}{3}}}{{{{10}^y} + {{10}^{ - y}}}}\]implies for\[y\].
If \[y = \dfrac{2}{3}\dfrac{{{{10}^x} - {{10}^{ - x}}}}{{{{10}^x} + {{10}^{ - x}}}}\]
\[{10^{2x}} = \dfrac{{3y + 2}}{{2 - 3y}}\]
 Or \[x = \dfrac{1}{2}{\log _{10}}\dfrac{{2 + 3y}}{{2 - 3y}}\]
Replace \[x\] with \[y\]:
\[x = \dfrac{2}{3} \cdot \dfrac{{{{10}^y} - {{10}^{ - y}}}}{{{{10}^y} + {{10}^{ - y}}}}\]
Solve for \[y\]:
\[x = \dfrac{2}{3}\dfrac{{{{10}^y} - {{10}^{ - y}}}}{{{{10}^y} + {{10}^{ - y}}}}\]
\[\therefore {f^{ - 1}}(x) = \dfrac{1}{2}{\log _{10}}\dfrac{{2 + 3x}}{{2 - 3x}}\]

So, option B is correct.

Note: Many students misunderstand the "one input → one output" defining feature of a function. This issue arises because implicitly-defined curves do have "inverses", they just may not be the graph of a factor. A common way to think about this is that an implicit curve provides an approximation for what would be obtained if you defined your input and output functions explicitly.