# 数学代写|离散数学作业代写Discrete Mathematics代考|GRAPH OF FUNCTION

#### Doug I. Jones

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## 数学代写|离散数学作业代写discrete mathematics代考|GRAPH OF FUNCTION

Let $f$ be a function from A to B, i.e. for every $x \in \mathrm{A}$ there exists unique $y \in \mathrm{B}$ such that $y=f(x)$. Further note that using the functional notation, $f$ can be expressed as
$$f={(x, f(x)): x \in \mathrm{A}} .$$
This representation is known as the graph of the function $f$. Consider the functions $f_1: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f_1(x)=x+1$; and $f_2: R \rightarrow{-2,2}$ defined by $f_2(x)=\left{\begin{array}{cc}-2 & \text { if } x>0 \ 2 & \text { if } x<0\end{array}\right.$
The graphs of above functions are given below.

Now consider the relations $f_1:[-4,4] \rightarrow[-4,4]$ defined by $\left[f_1(x)\right]^2=16-x^2 ; x \in[-4,4]$ and $f_2: \mathrm{R} \rightarrow \mathrm{R}$ defined by $\left[f_2(x)\right]^2=16 x ; x \in \mathrm{R}$. The graphs of above relations are given below.
These are nothing but a circle and parabola respectively. Where in figure $-1, y=f_1(x)$ and in figure $-2, y=f_2(x)$.

From the graph it is clear that for one value of $x$ in the domain set leads to two values in the range set. Hence these relations are not functions.

## 数学代写|离散数学作业代写discrete mathematics代考|COMPOSITION OF FUNCTIONS

Let $f$ be a function from the set $\mathrm{A}$ to the set $\mathrm{B}$ and $g$ be a function from the set $\mathrm{B}$ to the set $\mathrm{C}$. Then the composition of the functions $\mathrm{f}$ and $\mathrm{g}$ is given as $\left(g_0 f\right)$ or $g f$. This is a function from the set A to the set C. It may also be noted that domain of $g$ is equal to co-domain of $f$.

As $f$ is a function from the set A to the set B, then for every $x \in$ A there exists unique $y \in B$ such that $y=f(x)$. Similarly $g$ is a function from the set B to the set $\mathrm{C}$, then for every $y \in \mathrm{B}$ there exists unique $z \in \mathrm{C}$ such that $z=g(y)$. Again $\left(g_0 f\right)$ is a function from the set A to the set $\mathrm{C}$, so we get i.e. $\left(g_{\circ} f\right)(x)=z$ for all $x \in$ A.
i.e. $(g \circ f)(x)=g(y)$ $\left(g_{\circ} f\right)(x)=g(f(x))$
Consider two functions $f(x)=2 x+5$ and $g(x)=3 x$.
Therefore $\left(g_{\circ} f\right)(x)=g(f(x))$
i.e.
Similarly,
$=g(2 x+5)$
$=3(2 x+5)$
$\left(g_{\circ} f\right)(x)=6 x+15$
$\left(f_{\circ} g\right)(x)=f(g(x))$
$=f(3 x)$

i.e.
$$=2(3 x)+5$$
$$\left(f_{\mathrm{o}} g\right)(x)=6 x+5$$
4.5.1 Theorem
Let $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ be two functions. Then $\left(g_0 f\right)$ is one-one if both $f$ and $g$ are one-one and $\left(g_o f\right)$ is onto if both $f$ and $g$ are onto.

Proof: Let $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ be two functions. Since $f$ is a function from the set $\mathrm{A}$ to the set $\mathrm{B}$, then for every $x \in \mathrm{A}$ there exists unique $y \in \mathrm{B}$ such that $y=f(x)$. Similarly $g$ is a function from the set $\mathrm{B}$ to the set $\mathrm{C}$, then for every $y \in \mathrm{B}$ there exists unique $z \in \mathrm{C}$ such that $z=g(y)$.
Suppose the $f$ and $g$ are both one-one. Our claim is $\left(g_{\circ} f\right)$ is one-one. Since $f: \mathrm{A} \rightarrow \mathrm{B}$ and $g: \mathrm{B} \rightarrow \mathrm{C}$ we have $\left(g_{\circ} f\right): \mathrm{A} \rightarrow \mathrm{C}$.
Let $x_1, x_2 \in \mathrm{A}$ and $\left(g_{\circ} f\right)\left(x_1\right)=\left(g_{\circ} f\right)\left(x_2\right)$
This implies $\quad g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)$
i.e.
$f\left(x_1\right)=f\left(x_2\right)$
i.e.
$x_1=x_2$
$[\because g$ is one-one $]$
$[\because f$ is one-one $]$
Therefore $g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)$ implies $x_1=x_2$. So $\left(g_{\circ} f\right)$ is one-one.
Suppose that both $f$ and $g$ are onto. Since $g$ is onto, for every $z \in \mathrm{C}$ there is at least one $y \in \mathrm{B}$ such that $g(y)=z$. Again as $f$ is onto, for every $y \in \mathrm{B}$ there exists at least one $x \in$ A such that $f(x)=y$.

As a result for every $z \in \mathrm{C}$ there is at least one $x \in \mathrm{A}$ such that $\left(g_{\circ} f\right)(x)=z$. Therefore $\left(g{ }_{\circ} f\right)$ is onto.

# 离散数学代写

## 数学代写|离散数学作业代写discrete mathematics代考|GRAPH OF FUNCTION

$$f={(x, f(x)): x \in \mathrm{A}} .$$

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