Question

How do you find the inverse of \[f(x) = 2 - 2{x^2}?\]

Answer 1

Hint: We will equate the above function with a variable as inverse of that variable gives us the same function.
Then we will find the value of x from the equated equation. Later we will get the inverse function as x will be equal to inverse function in y.

Formula used: If a function \[f(x) = y\], then it implies the inverse function also:
\[{f^{ - 1}}(y) = x\] .

Complete step-by-step solution:
Let's say the above function is defined as \[f(x) = y\].
Then the inverse of the function would be \[{f^{ - 1}}(y) = x\].
But it is given that, \[f(x) = 2 - 2{x^2}\]
So, it should be \[y = f(x) = 2 - 2{x^2}\]
Now, subtracting \[2\] from both the side, we get the following equation:
\[ \Rightarrow y - 2 = 2 - 2{x^2} - 2\]
Or, \[y - 2 = - 2{x^2}\]
Now multiply both the sides by \[ - 1\], we get:
\[ \Rightarrow 2 - y = 2{x^2}\].
Now divide both the sides by \[2\], we get:
\[ \Rightarrow \dfrac{{2 - y}}{2} = {x^2}\]
Now, taking the square root on the both sides, we get following equation:
\[ \Rightarrow \sqrt {\dfrac{{2 - y}}{2}} = x\]
And, now if we tally with the above equation, we can derive the following equation:
\[ \Rightarrow x = {f^{ - 1}}(y) = \sqrt {\dfrac{{2 - y}}{2}} \].
If we replace the value of \[y\] by \[x\] then we can say that:
\[ \Rightarrow {f^{ - 1}}(x) = \sqrt {\dfrac{{2 - x}}{2}} \].

\[\therefore \]The inverse of \[f(x) = 2 - 2{x^2}\] is \[\sqrt {\dfrac{{2 - x}}{2}} \].

Note: The inverse of a function is a function that is reverse or reciprocal of that function.
If the function \[f\] applied to an input \[x\] gives a result of \[y\], then applying its inverse function \[g\] to \[y\] will give us the result of \[x\].
Always remember that the inverse of a function is denoted by \[{f^{ - 1}}\].
Some properties of a function:
There is an always symmetry relationship exist between function and its inverse, that is why it states:
\[{\left( {{f^{ - 1}}} \right)^{ - 1}} = f\]
If an inverse function exists for a given function then it must be unique by its property.