How do you know if $y=2x+7$ is a function? \[\]

Hint: We recall the definition of a function, domain and range of a function. If $X$ is a domain and $Y$ is the range then the relation $f:X\to Y$ is a function if for every $x\in X$ there is a $y\in Y$ and if $\left( {{x}_{1}},{{y}_{1}} \right)\in f,\left( {{x}_{2}},{{y}_{2}} \right)\in f$ and ${{x}_{1}}={{x}_{2}}$then ${{y}_{1}}={{y}_{2}}$. We take $y=f\left( x \right)=2x+1$ as the relation and test using these conditions whether $f$ is a function or not. \[\]

Complete step-by-step answer:
We know that a function or map is a relation which relates inputs to outputs. The set from which the functions takes inputs is called domain and the set from which functions returns outputs is called co-domain. The set of only outputs which is a subset of co-domain is called range. If the relation $f:X\to Y$ is a function it has to satisfy 2 conditions, \[\]
1. All elements or inputs of the domain set are mapped which means for every $x\in X$ there is $y\in Y$. \[\]
2. One inputs cannot have two outputs. It means for all $\left( {{x}_{1}},{{y}_{1}} \right)\in f,\left( {{x}_{1}},{{y}_{2}} \right)\in f$ and ${{x}_{1}}={{x}_{2}}$then ${{y}_{1}}={{y}_{2}}$ \[\]
We are given the relation between $x,y$ as $y=2x+1$. We can take real number set $\mathsf{\mathbb{R}}$ for input $x$. We see that for every real number $x$ as inputs; we can get a real number $y$ which returns twice $x$ plus 1 as outputs. So for every $x\in \mathsf{\mathbb{R}}$ there is an $y\in \mathsf{\mathbb{R}}$. So our first condition is satisfied. \[\]
Let us have $\left( {{x}_{1}},{{y}_{1}} \right)\in f,\left( {{x}_{1}},{{y}_{2}} \right)\in f$ where ${{x}_{1}},{{x}_{2}}\in \mathsf{\mathbb{R}}$ and ${{y}_{1}},{{y}_{2}}\in \mathsf{\mathbb{R}}$. If ${{x}_{1}}={{x}_{2}}$then we have
\[\begin{align}
  & {{x}_{1}}={{x}_{2}} \\
 & \Rightarrow 2{{x}_{1}}=2{{x}_{2}} \\
 & \Rightarrow 2{{x}_{1}}+1=2{{x}_{2}}+1 \\
\end{align}\]
We can assign $2{{x}_{1}}+1={{y}_{1}},2{{x}_{2}}+2={{y}_{2}}$ to have ${{y}_{1}}={{y}_{2}}$, Hence one input cannot have two outputs . Our second condition is satisfied. So $y=2x+1$ is a function. \[\]

Note: We note that if for every output $y\in Y$ there is an input $x\in X$ then $f:X\to Y$ is onto a function where range and codomain are the same. If for every $y\in Y$ there is exactly one $x\in X$ then $f:X\to Y$ is one-one. If $f$ is one-one and onto it has an inverse function ${{f}^{-1}}:Y\to X$. Here in this problem the function $f:\mathsf{\mathbb{R}}\to \mathsf{\mathbb{R}},f\left( x \right)=2x+1$ is one-one and onto has inverse function ${{f}^{-1}}:\mathsf{\mathbb{R}}\to \mathsf{\mathbb{R}},{{f}^{-1}}\left( x \right)=\dfrac{x-1}{2}$.