
Let $\mathbb{N}$ be the set of natural numbers. Define a real-valued function $f:\mathbb{N}\to \mathbb{N}$ by $f\left( x \right)=2x+1$. Using the definition, complete the table.
x 1 2 3 4 5 6 7 y $f\left( 1 \right)$ $f\left( 2 \right)$ $f\left( 3 \right)$ $f\left( 4 \right)$ $f\left( 5 \right)$ \[f\left( 6 \right)\] $f\left( 7 \right)$
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| y | $f\left( 1 \right)$ | $f\left( 2 \right)$ | $f\left( 3 \right)$ | $f\left( 4 \right)$ | $f\left( 5 \right)$ | \[f\left( 6 \right)\] | $f\left( 7 \right)$ |
Answer
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Hint: We first explain the function and its image values for the given inputs. We get all the input points and by putting the value we get the outputs as image points. We complete the table.
Complete step by step solution:
We have been given a table and the conditions where the given real-valued function for $f:\mathbb{N}\to \mathbb{N}$ is $f\left( x \right)=2x+1$. Let $\mathbb{N}$ be the set of natural numbers.
We have to find the values for $y=f\left( x \right),x=1 to 7$.
The basic requirement for a function to have input and evaluate its output. The input is called the domain of the function and the outcome is called the range.
We generally define a function as $f\left( x \right)$. We can also express it as $f:x\to f\left( x \right)$.
We put the values of $x$ in the given equation to find the values of $y$.
So, putting 1 we get, $f\left( 1 \right)=2\times 1+1=3$, putting 2 we get, $f\left( 2 \right)=2\times 2+1=5$.
Putting 3 we get, $f\left( 3 \right)=2\times 3+1=7$, putting 4 we get, $f\left( 4 \right)=2\times 4+1=9$.
Putting 5 we get, $f\left( 5 \right)=2\times 5+1=11$, putting 6 we get, $f\left( 6 \right)=2\times 6+1=13$.
Putting 7 we get, $f\left( 7 \right)=2\times 7+1=15$.
Now we complete the table and get
Note: The difference between range and image is that image points are the subset of the whole possible outcomes. Image points are the particular outcomes of the given inputs. Thus, the values 3, 5, 11 are the elements of $f\left( x \right),x\in \mathbb{N}$.
Complete step by step solution:
We have been given a table and the conditions where the given real-valued function for $f:\mathbb{N}\to \mathbb{N}$ is $f\left( x \right)=2x+1$. Let $\mathbb{N}$ be the set of natural numbers.
We have to find the values for $y=f\left( x \right),x=1 to 7$.
The basic requirement for a function to have input and evaluate its output. The input is called the domain of the function and the outcome is called the range.
We generally define a function as $f\left( x \right)$. We can also express it as $f:x\to f\left( x \right)$.
We put the values of $x$ in the given equation to find the values of $y$.
So, putting 1 we get, $f\left( 1 \right)=2\times 1+1=3$, putting 2 we get, $f\left( 2 \right)=2\times 2+1=5$.
Putting 3 we get, $f\left( 3 \right)=2\times 3+1=7$, putting 4 we get, $f\left( 4 \right)=2\times 4+1=9$.
Putting 5 we get, $f\left( 5 \right)=2\times 5+1=11$, putting 6 we get, $f\left( 6 \right)=2\times 6+1=13$.
Putting 7 we get, $f\left( 7 \right)=2\times 7+1=15$.
Now we complete the table and get
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| $y=f\left( x \right)$ | 3 | 5 | 7 | 9 | 11 | 13 | 15 |
Note: The difference between range and image is that image points are the subset of the whole possible outcomes. Image points are the particular outcomes of the given inputs. Thus, the values 3, 5, 11 are the elements of $f\left( x \right),x\in \mathbb{N}$.
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