
How do you find the inverse of $f(x) = {x^2} - 2x - 8$ and is it a function?
Answer
516.6k+ 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.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

