The inverse function of $f\left(x\right) = \frac{8^{2x}-8^{-2x}}{8^{2x} + 8^{-2x}}, x\epsilon \left(-1, 1\right),$ is __________.
- $\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1-x}{1+x}\right)$
- $\frac{1}{4}\left(log_{8} \,e\right) log_{e} \left(\frac{1+x}{1-x}\right)$
- $\frac{1}{4} log_{e} \left(\frac{1+x}{1-x}\right)$
- $\frac{1}{4} log_{e} \left(\frac{1- x}{1+ x}\right)$
The Correct Option is B
Solution and Explanation
so, $8^{4x}+1=\frac{2}{1-y} \Rightarrow 8^{4x}=\frac{1+y}{1-y}$
$\Rightarrow x=\ell n\left(\frac{1+y}{1-y}\right)\times \frac{1}{4\ell n8}=f ^{-1}\left(y\right)$
Hence, $f ^{-1}\left(x\right)=\frac{1}{4}log_{g}\,e\ell n\left(\frac{1+x}{1-x}\right)$
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions