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Question

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

How do you find the inverse of f(x)=3x5?f\left( x \right)=3x-5?

Explanation

Solution

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 xx. After that we will get the obtained value of xx in the function ff and equate it with yy then multiplying both sides by f1{{f}^{1}} we will get our required answer.

Complete step by step solution:
In this question we have given the function f(x)=3x5f\left( x \right)=3x-5 .
In order to find inverse of given function, Let us consider, y=f(x);yy=f\left( x \right);y
Where yy is arbitrarily chosen.
Which implies: y=3x5y=3x-5
Now we will add 55 on both the sides which will balance the equation.
3x=y+5\Rightarrow 3x=y+5
Now we will rearrange the above equation as follows. 3x=y+53x=y+5
Now, we will divide both sides of above equation by 33 and we get,
As we have consider earlier x=y+53...(i)\Rightarrow x=\dfrac{y+5}{3}...(i)
Now, we will substitute the value of xx which we have obtain in equation (1)\left( 1 \right) in equation y=f(x)y=f\left( x \right)
Therefore, we have y=f(y+53)y=f\left( \dfrac{y+5}{3} \right)
Now, we will multiply both the sides of above equation by f1{{f}^{-1}} and we get,
f1(y)=(y+53).{{f}^{-1}}\left( y \right)=\left( \dfrac{y+5}{3} \right). (Since we know that AA1=1A{{A}^{-1}}=1)
Since, the choice of the variable is made arbitrary therefore we can rewritten this as
f(x)=3x5f(x)=3x-5
Hence, The increase of f(x)=x+53f(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 ff takes aa to bb then, the inverse of f'f' must be taken bb to aa.
If we have a function as for FF then the inverse function is devoted by f1{{f}^{1}} or F1{{F}^{1}} respectively.
That is, afbf1aa\to f\to b\to {{f}^{-1}}\to a or in other words we can write it as f(a)=bf1(b)=af(a)=b\Leftrightarrow {{f}^{-1}}(b)=a.