How do you find the inverse of $f(x) = {x^2} - 2x - 8$ and is it a function?
Answer
598.2k+ views
Hint: To find the inverse, equate the function to another variable and then find the value of the pre-image from this image. Also, check if the inverse is really a function, by using the well-defined property of functions.
Complete step by step solution:
Inverse function, also known as an anti-function, is defined as a function which will take the result of a function and give back the original input of the function. In other words, it takes the image of a function and gives back its preimage.
Since, there is no domain and codomain of $f$ is mentioned, for simplicity let us assume the entire real line $\mathbb{R}$ as both the domain and co-domain of the function.
Let us take the given function \[f(x) = y\].
$ \Rightarrow y = f(x) = {x^2} - 2x - 8$. - - - - - - - - - - - - - - - - - - - - - (1)
Then the inverse of the function would be ${f^{ - 1}}(y) = x$. - - - - - - - - - - - - - - - (2)
Now, we will simplify the equation (1) to get a function which gives back x. For this express the equation in the form $x = g(y)$. Here, $g(y)$ will be the required inverse.
From equation (1) we have,
$y = {x^2} - 2x - 8$
Now, observe that we can use the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ as we have both the $x{\text{ and }}{x^2}$ terms in the equation. So, for that we will recombine terms slightly.
$ \Rightarrow y = {x^2} - 2x + 1 - 9$
Now, on adding 9 on both sides we get,
$ \Rightarrow y + 9 = {x^2} - 2x + 1$
Now, use the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to get,
$ \Rightarrow y + 9 = {\left( {x - 1} \right)^2}$
Now, taking root on both sides we get,
$ \Rightarrow \pm \sqrt {y + 9} = \left( {x - 1} \right)$
$ \Rightarrow \pm \sqrt {y + 9} + 1 = x$
$ \Rightarrow x = 1 \pm \sqrt {y + 9} $
$ \Rightarrow {f^{ - 1}}(y) = 1 \pm \sqrt {y + 9} $ [From (2)]
On replacing y with x,
$ \Rightarrow {f^{ - 1}}(x) = 1 \pm \sqrt {x + 9} $
And $g(x) = 1 \pm \sqrt {x + 9} $, is the inverse of the function $f(x) = {x^2} - 2x - 8$.
But this $g(x) = 1 \pm \sqrt {x + 9} $ is not a function, because one can get two different values $4{\text{ and }} - 2$ for $x = 0$. And this violates the well-defined property of a function, which states that ${x_1} = {x_2} \Rightarrow f({x_1}) = f({x_2})$ for every x in the domain.
Note: Note that, inverse is not a function always. So, in order to make it a function we have to look at the domain and range of the function and try to make its inverse a function by modifying the domain of the inverse slightly.
Complete step by step solution:
Inverse function, also known as an anti-function, is defined as a function which will take the result of a function and give back the original input of the function. In other words, it takes the image of a function and gives back its preimage.
Since, there is no domain and codomain of $f$ is mentioned, for simplicity let us assume the entire real line $\mathbb{R}$ as both the domain and co-domain of the function.
Let us take the given function \[f(x) = y\].
$ \Rightarrow y = f(x) = {x^2} - 2x - 8$. - - - - - - - - - - - - - - - - - - - - - (1)
Then the inverse of the function would be ${f^{ - 1}}(y) = x$. - - - - - - - - - - - - - - - (2)
Now, we will simplify the equation (1) to get a function which gives back x. For this express the equation in the form $x = g(y)$. Here, $g(y)$ will be the required inverse.
From equation (1) we have,
$y = {x^2} - 2x - 8$
Now, observe that we can use the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ as we have both the $x{\text{ and }}{x^2}$ terms in the equation. So, for that we will recombine terms slightly.
$ \Rightarrow y = {x^2} - 2x + 1 - 9$
Now, on adding 9 on both sides we get,
$ \Rightarrow y + 9 = {x^2} - 2x + 1$
Now, use the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to get,
$ \Rightarrow y + 9 = {\left( {x - 1} \right)^2}$
Now, taking root on both sides we get,
$ \Rightarrow \pm \sqrt {y + 9} = \left( {x - 1} \right)$
$ \Rightarrow \pm \sqrt {y + 9} + 1 = x$
$ \Rightarrow x = 1 \pm \sqrt {y + 9} $
$ \Rightarrow {f^{ - 1}}(y) = 1 \pm \sqrt {y + 9} $ [From (2)]
On replacing y with x,
$ \Rightarrow {f^{ - 1}}(x) = 1 \pm \sqrt {x + 9} $
And $g(x) = 1 \pm \sqrt {x + 9} $, is the inverse of the function $f(x) = {x^2} - 2x - 8$.
But this $g(x) = 1 \pm \sqrt {x + 9} $ is not a function, because one can get two different values $4{\text{ and }} - 2$ for $x = 0$. And this violates the well-defined property of a function, which states that ${x_1} = {x_2} \Rightarrow f({x_1}) = f({x_2})$ for every x in the domain.
Note: Note that, inverse is not a function always. So, in order to make it a function we have to look at the domain and range of the function and try to make its inverse a function by modifying the domain of the inverse slightly.
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