How do you find $\left( h-g \right)\left( t \right)=h\left( t \right)-g\left( t \right)$ . Given $h\left( t \right)=2t+1$ and $g\left( t \right)=2t+2$?

Hint: Now we are given that $h\left( t \right)=2t+1$ and $g\left( t \right)=2t+2$ . To find the value of $\left( h-g \right)\left( t \right)$ we will have to subtract the two functions. Hence we will consider $h\left( t \right)-g\left( t \right)$ . Now we will simplify the obtained equation and hence find the value of $\left( h-g \right)\left( t \right)$ .

Complete step by step solution:
Now consider the given functions $h\left( t \right)=2t+1$ and $g\left( t \right)=2t+2$
Now we know that the given functions are linear functions in one variable where the variable is t.
Now we want to find the value of $\left( h-g \right)\left( t \right)$ . Now $\left( h-g \right)\left( t \right)$ is a nothing but the function $\left( h-g \right)$ in t.
Now to find this function is nothing but just subtraction of two given functions.
Hence we have,
$\left( h-g \right)\left( t \right)=h\left( t \right)-g\left( t \right)$
Now substituting the value of function h and function g we get,
$\begin{align}
  & \Rightarrow \left( h-g \right)\left( t \right)=2t+1-\left( 2t+2 \right) \\
 & \Rightarrow \left( h-g \right)\left( t \right)=2t-2t-1 \\
\end{align}$
Now we know that we can add and subtract the terms with the same variables and same degree, Hence we can say $2t-2t=0$ . Using this we get,
$\Rightarrow \left( h-g \right)\left( t \right)=-1$

Hence the value of the function $\left( h-g \right)\left( t \right)$ is - 1. Also we can see that the function obtained is a constant function independent from variable t.

Note: Now note that we can create a new function out of two functions by adding and subtracting the function. We also have a composite function written as $f\left( g\left( x \right) \right)$ . To create a composite function we substitute the value of x as a function. For example $f\left( x \right)={{x}^{2}}$ and $g\left( x \right)=2x+2$ then to write the function $f\left( g\left( x \right) \right)$ we will substitute the value of $x=g\left( x \right)$ in $f\left( x \right)$ . Hence we get, $f\left( g\left( x \right) \right)={{\left( 2x+2 \right)}^{2}}$ similarly $g\left( f\left( x \right) \right)=2{{x}^{2}}+2$ .