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# Let $f:R\to R:f\left( x \right)=10x+3$. Find ${{f}^{-1}}$.

Last updated date: 20th Jun 2024
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Hint: In this problem we have given the function $f\left( x \right)$ as $10x+3$, and asked to calculate the value of ${{f}^{-1}}$. For this we will first take the assumption $y=10x+3$ and solve for the $x$. To solve the above equation, we will apply the reverse athematic operations for the operations we have in the equation. We can observe that the addition and multiplication operations are involved in the given equation. So, we will apply reverse arithmetic operations for addition and multiplication which are subtraction and division. After applying all the arithmetic operations, we will get the value of $x$ which is our required value.

Complete step by step solution:
Given that, $f:R\to R:f\left( x \right)=10x+3$.
Let us assume $y=10x+3$. To find the value of ${{f}^{-1}}$ we need to have the value of $x$ from the above equation.
In the above equation we can observe that $3$ is in addition, so we will subtract $3$ from both sides of the above equation, then we will get
$\Rightarrow y-3=10x+3-3$
We know that $+a-a=0$, then the above equation is modified as
$\Rightarrow y-3=10x$
In the above equation we can observe that $10$ is in multiplication, so we will divide with $10$ on both sides of the above equation, then we will get
$\Rightarrow \dfrac{y-3}{10}=\dfrac{10x}{10}$
We know that $\dfrac{a}{a}=1$, then we will have
$\Rightarrow \dfrac{y-3}{10}=x$

Here we have the value of $x$ as $\dfrac{y-3}{10}$. Hence the value of ${{f}^{'}}\left( x \right)$ is $\dfrac{x-3}{10}$.

Note: We can check whether the obtained answer is correct or wrong by using the relation between the function $f\left( x \right)$, ${{f}^{'}}\left( x \right)$ which is “if $f\left( b \right)=a$ then ${{f}^{-1}}\left( a \right)$ must be $b$”. We can also have the relation ${{f}^{-1}}\left( f\left( a \right) \right)=a$. From these relations we can check if the result is correct or wrong.