Question

Let $f:N\to R$ be a function defined as $f\left( x \right)=4{{x}^{2}}+12x+15$. Show that the function $f:N\to \text{Range}\left( f \right)$ is invertible. Also find the inverse of $f$.

Answer 1

Hint: We prove that the function $f$ is invertible by proving $f$ is onto and one-one. We prove $f$ is one-one by proving $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}={{x}_{2}}$for some${{x}_{1}},{{x}_{2}}\in N$. We prove $f$ is onto by proving the co-domain and range of the function $f$ is the same. We find the inverse using the property $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and then using quadratic formula. \[\]

Complete step by step answer:
We know that if a function is invertible then it is one-one and onto. One-one implies for every element in the domain called preimage of the function we are going to get exactly image in the range. The set of all images are called range and any superset of range is called co-domain. Onto implies every element in the co-domain set has a pre-image that means co-domain is the range. We denote any function with domain $A$ and co-domain $B$ as
\[f:A\to B\]
We are given the function in the question as,
\[f\left( x \right)=4{{x}^{2}}+12x+15\]
Here we are also given the question $f:N\to R$ and we are asked to prove $f:N\to \text{Range}\left( f \right)$ is invertible. Here in the statement of proof the domain set is a set of natural numbers. \[\]
Here we have to prove $f:N\to \text{Range}\left( f \right)$ is one-one and onto. We take two elements from the domain set $N$ say ${{x}_{1}},{{x}_{2}}$ and then denote their images as $f\left( {{x}_{1}} \right),f\left( {{x}_{2}} \right)$. If we can prove ${{x}_{1}}={{x}_{2}}$whenever $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$we get unique pre-images which means then $f$ is one-one. We have
\[\begin{align}
  & f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right) \\
 & \Rightarrow 4{{x}_{1}}^{2}+12{{x}_{1}}+15=4{{x}_{2}}^{2}+12{{x}_{2}}+15 \\
 & \Rightarrow 4\left( {{x}_{1}}^{2}-{{x}_{2}}^{2} \right)+12\left( {{x}_{1}}-{{x}_{2}} \right)=0 \\
 & \Rightarrow \left( {{x}_{1}}-{{x}_{2}} \right)\left( {{x}_{1}}+{{x}_{2}} \right)+3\left( {{x}_{1}}-{{x}_{2}} \right)=0 \\
 & \Rightarrow \left( {{x}_{1}}-{{x}_{2}} \right)\left( {{x}_{1}}+{{x}_{2}}+3 \right)=0 \\
\end{align}\]
So we either have ${{x}_{1}}-{{x}_{2}}=0$ which implies ${{x}_{1}}={{x}_{2}}$or we have ${{x}_{1}}+{{x}_{2}}+3=0$ which implies ${{x}_{1}}+{{x}_{2}}=-3$ but have taken ${{x}_{1}},{{x}_{2}}$ from the set of natural number which means they are positive integers and their sum cannot be negative integer like $-3$. So we reject the possibility ${{x}_{1}}+{{x}_{2}}+3=0$ and conclude${{x}_{1}}={{x}_{2}}$. So $f$ is one-one.
We see in the statement of the proof “$f:N\to \text{Range}\left( f \right)$ is invertible” that co-domain$\left( f \right)$ is also the range$\left( f \right)$. So $f$ is onto. \[\]
We $f$ as one-one and onto and hence $f$ is invertible. Let us have the dependent variable $y$ assigned to the function. Let us denote the inverse of $f$ as ${{f}^{-1}}$ where ${{f}^{-1}}:R\text{ange}\left( f \right)\to N$. We have from property of inverse function
\[\begin{align}
  & f\left( {{f}^{-1}}\left( x \right) \right)=x \\
 & \Rightarrow 4{{\left( {{f}^{-1}}\left( x \right) \right)}^{2}}+12{{f}^{-1}}\left( x \right)+15=x \\
 & \Rightarrow 4{{\left( {{f}^{-1}}\left( x \right) \right)}^{2}}+12{{f}^{-1}}\left( x \right)+15-x=0 \\
\end{align}\]
Let us use the quadratic formula for the root of quadratic function and have,
\[\begin{align}
  & \Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{-12\pm \sqrt{{{12}^{2}}-4\times 4\times \left( 15-x \right)}}{2\times 4} \\
 & \Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{-12\pm \sqrt{16x-96}}{8} \\
 & \Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{-12\pm 4\sqrt{x-6}}{8} \\
 & \Rightarrow {{f}^{-1}}\left( x \right)=\dfrac{-3\pm \sqrt{x-6}}{2} \\
\end{align}\]
 Since ${{f}^{-1}}\left( x \right)\in N $for all $x\in \text{Range}\left( f \right)$ so ${{f}^{-1}}\left( x \right)$ will be positive, we reject ${{f}^{-1}}\left( x \right)=\dfrac{-3-\sqrt{x-6}}{2}$

hence the inverse function is- \[{{f}^{-1}}\left( x \right)=\dfrac{-3+\sqrt{x-6}}{2}\]

Note: We can verify the result by checking ${{f}^{-1}}\left( f\left( x \right) \right)=x$. When a function is both one-one and onto it is also called bijective. The inverse function and the original function are always symmetrical about the line $y=x$. The graph is not an upward opened parabola because the domain is $N$.