
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 $g\left( t \right)=2t+2$?
Answer
538.8k+ views
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 $g\left( f\left( x \right) \right)=2{{x}^{2}}+2$ .
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 $g\left( f\left( x \right) \right)=2{{x}^{2}}+2$ .
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

India is a sovereign socialist secular democratic republic class 12 social science CBSE

How many states of matter are there in total class 12 chemistry CBSE

What are the advantages of vegetative propagation class 12 biology CBSE

Suicide bags of cells are aEndoplasmic reticulum bLysosome class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

