Answer to Question #230547 in Discrete Mathematics for Mak Kin Yun Riva

Question

5a) Explain whether each of the following relations on the set of real numbers is a function or not. For those (if any) that are indeed functions say whether they are one-to-one and/or onto. (2 marks)

i) y = f(x) = 2x2+1 xэR, y эR

ii) y = g(x) = 1/(x+1) (xэR, y эR , x != -1)

iii) Let h be a function from X = {1, 2, 3, 4} to Y = {a, b, c, d}.

h(1) = d, h(1) = c, h(2) = a, h(3) =b, and h(4) = b.

5b) Does either f or g have an inverse? If so, find this inverse. (1 marks)

5c) Find the composite functions f 。g and g。f . (2marks)

Question 6

Answer 1

5a) Let us explain whether each of the following relations on the set of real numbers is a function or not. For those (if any) that are indeed functions let us say whether they are one-to-one and/or onto.

i) The relation $y = f(x) = 2x^2+1,\ x\in R, y \in R$ is a function because of for each $x\in \R$ there exists a unique $y\in\R$ such that $y=2x^2+1=f(x).$ Since $f(-1)=2(-1)^2+1=2\cdot 1^2+1=f(1),$ this function is not one-to-one. Taking into account that the equation $0=2x^2+1$ has no real solution, we conclude that this function is not onto $\R$.

ii) The relation $y = g(x) = \frac{1}{x+1} (x\in \R, y \in\R , x \ne -1)$ is a function because of for each $x\in \R\setminus\{-1\}$ there exists a unique $y\in\R$ such that $y= \frac{1}{x+1}=g(x).$ Since $g(a)=g(b)$ implies $\frac{1}{a+1}=\frac{1}{b+1},$ and hence $a+1=b+1,$ we conclude that $a=b,$ and thus this function is not one-to-one. Taking into account that the equation $0=\frac{1}{x+1}$ has no real solution, we conclude that this function is not onto $\R$.

iii) Let $h$ be a function from $X =\{1, 2, 3, 4\}$ to $Y = \{a, b, c, d\}$.

$h(1) = d,\ h(1) = c,\ h(2) = a,\ h(3) =b,$ and $h(4) = b.$

This relation is not a function because of for $x=1$ there exist two values $y=c$ and $y=d$ such that $y=h(x).$

5b) Since $f:\R\to \R$ is not a bijection, it has no inverse function $f^{-1}:\R\to \R$.

Since $g:\R\setminus\{-1\}\to \R$ is not onto, it has no inverse function $g^{-1}:\R\to \R\setminus\{-1\}.$

Taking into account that $f$ is one-to-one and for any $y\ne 0$ there exists a unique $x=\frac{1}{y}-1$ such that $g(x)=\frac{1}{(\frac{1}{y}-1)+1}=y,$ we concvlude that there exist an inverse function $g^{-1}:\R\setminus\{0\}\to \R\setminus\{-1\}, \ g^{-1}(x)=\frac{1}{x}-1.$

5c) Let us find the composite functions $f\circ g$ and $g\circ f:$

$f\circ g(x)=f(g(x))=f(\frac{1}{x+1})=2(\frac{1}{x+1})^2+1=\frac{2}{(x+1)^2}+1;$

$g\circ f(x)=g(f(x))=g(2x^2+1)=\frac{1}{2x^2+1+1}=\frac{1}{2x^2+2}.$