Question

How do you find the inverse of \[f\left( x \right)=3x-5?\]

Answer 1

Hint: In order to find the inverse of a given function firstly we will rename our given function. After renaming the given function we will do all the required operations which can be additional subtraction multiplication or division in order to get the value of \[x\]. After that we will get the obtained value of \[x\] in the function \[f\] and equate it with \[y\] then multiplying both sides by \[{{f}^{1}}\] we will get our required answer.

Complete step by step solution:
In this question we have given the function \[f\left( x \right)=3x-5\] .
In order to find inverse of given function, Let us consider, $y=f\left( x \right);y$
Where \[y\] is arbitrarily chosen.
Which implies: $y=3x-5$
Now we will add \[5\] on both the sides which will balance the equation.
$\Rightarrow 3x=y+5$
Now we will rearrange the above equation as follows. \[3x=y+5\]
Now, we will divide both sides of above equation by \[3\] and we get,
As we have consider earlier $\Rightarrow x=\dfrac{y+5}{3}...(i)$
Now, we will substitute the value of \[x\] which we have obtain in equation \[\left( 1 \right)\] in equation \[y=f\left( x \right)\]
Therefore, we have \[y=f\left( \dfrac{y+5}{3} \right)\]
Now, we will multiply both the sides of above equation by \[{{f}^{-1}}\] and we get,
\[{{f}^{-1}}\left( y \right)=\left( \dfrac{y+5}{3} \right).\] (Since we know that \[A{{A}^{-1}}=1\])
Since, the choice of the variable is made arbitrary therefore we can rewritten this as
$f(x)=3x-5$
Hence, The increase of $f(x)=\dfrac{x+5}{3}$

Note: An inverse function or an anti-function is defined as a function, which can reverse into another function.
In simple words we can say that, If \[f\] takes \[a\] to \[b\] then, the inverse of \['f'\] must be taken \[b\] to \[a\].
If we have a function as for \[F\] then the inverse function is devoted by \[{{f}^{1}}\] or \[{{F}^{1}}\] respectively.
That is, \[a\to f\to b\to {{f}^{-1}}\to a\] or in other words we can write it as $f(a)=b\Leftrightarrow {{f}^{-1}}(b)=a$.