Question

Inverse of the function \[f:R \to ( - \infty ,1)\] given by \[f(x) = 1 - {2^{ - x}}\]
A. \[ - {\log _2}(1 - x)\]
B. \[ - {\log _2}(x)\]
C. \[0\]
D. \[1\]

Answer 1

Hint: To solve this question first, we have to replace \[f\left( x \right)\] with \[y\] and then we have to replace every x with a y and replace every \[y\] with an \[x\].

Complete step by step solution: A mirror image or reversal of the original function is what is meant by the phrase "inverse of the function." In other words, if you know the inverse, you can use it to solve any equation that uses that function.
In general, it is simpler to invert a function when it is expressed in a single variable form than when it is expressed in a number of variables. But there are some techniques that can make inverting functions simpler.
Inverse function is nothing but the inverse of a given function. In this particular case, we are looking for the inverse of \[f:R \to ( - \infty ,1)\]given by \[f\left( x \right) = 1 - 2 - x\].
We have the question \[f(x) = 1 - {2^{ - x}}\]
Let \[y = 1 - {2^{ - x}}\] or \[{2^{ - x}} = 1 - y\]
Or
\[ - x = {\log _2}(1 - y)\]
Or
\[{f^{ - 1}}(x) = g(x)\]
Or
\[ = - {\log _2}(1 - x)\]
The inverse function can be found by setting the derivative of \[f\] to \[0\] and solving for \[x\]. This process is known as differentiation with respect to x or quotienting. The derivative of \[f\] at \[x = 0\] is \[ - 1\], which means that the inverse function takes on the value \[ - lo{g_2}\left( {1 - x} \right)\].

So, option A is correct.

Note: Even if students are following the instructions set \[f(x) = y\], swap all instances \[x \leftrightarrow y\], they often will correctly set up. And then, unfortunately, we have no idea how to start solving for \[y\] in terms of\[x\].
If inverse\[\left( x \right) = - 2 - {x^2} + lo{g_2}\left( {1 - x} \right) + C\]; where \[C\] is a constant. While it may seem simple at first, there are certain things that students can fail to take into account when trying to solve this equation. One of the most common mistakes made is failing to differentiate between real and imaginary parts of the equation. When working with complex math problems like this, it's important to be patient and understand all of the relationships involved before reaching a solution.