Let $f$ and $g$ be real functions defined by $f(x)=2 x+1$ and $g(x)=4 x-7$.
(i) For what real numbers $x, f(x)=g(x)$ ?
(ii) For what real numbers $x, f(x)< g(x)$ ?
We have, $$f(x)=2 x+1 \text { and } g(x)=4 x-7$$
(i) $\because$ $$f(x)=g(x)$$
$$\begin{aligned} & \Rightarrow \quad 2 x+1=4 x-7 \Rightarrow 2 x=8 \\ & \therefore \quad x=4 \end{aligned}$$
(ii) $\because\quad f(x)< g(x)$
$\begin{array}{lc}\Rightarrow & 2 x+1<4 x-7 \\ \Rightarrow & 2 x-4 x+1<4 x-7-4 x \\ \Rightarrow & -2 x+1<-7 \\ \Rightarrow & -2 x<-7-1 \\ \Rightarrow & -2 x<-8 \\ \Rightarrow & \frac{-2 x}{-2}>\frac{-8}{-2} \\ \therefore & x>4\end{array}$
If $f$ and $g$ are two real valued functions defined as $f(x)=2 x+1$ and $g(x)=x^2+1$, then find
(i) $f+g$
(ii) $f-g$
(iii) $f g$
(iv) $\frac{f}{g}$
We have, $f(x)=2 x+1$ and $g(x)=x^2+1$
$$\begin{aligned} \text{(i)}\quad (f+g)(x) & =f(x)+g(x) \\ & =2 x+1+x^2+1=x^2+2 x+2 \end{aligned}$$
$$\begin{aligned} \text{(i)}\quad (f-g)(x) & =f(x)-g(x)=(2 x+1)-\left(x^2+1\right) \\ & =2 x+1-x^2-1=2 x-x^2=x(2-x) \end{aligned}$$
$$\begin{aligned} \text{(iii)}\quad (f g)(x) & =f(x) \cdot g(x)=(2 x+1)\left(x^2+1\right) \\ & =2 x^3+2 x+x^2+1=2 x^3+x^2+2 x+1 \end{aligned}$$
(iv) $\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{2 x+1}{x^2+1}$
Express the following functions as set of ordered pairs and determine their range.
$$f: x \rightarrow R, f(x)=x^3+1 \text {, where } x=\{-1,0,3,9,7\}$$
We have, $f: X \rightarrow R, f(x)=x^3+1$.
Where $X=\{-1,0,3,9,7\}$,
When $x=-1$, then $f(-1)=(-1)^3+1=-1+1=0$
$\begin{aligned} & x=0, \text { then } f(0)=(0)^3+1=0+1=1 \\ & x=3 \text { then } f(3)=(3)^3+1=27+1=28 \\ & x=9 \text {, then } f(9)=(9)^3+1=729+1=730 \\ & x=7, \text { then } f(7)=(7)^3+1=343+1=344 \\ & f=\{(-1,0),(0,1),(3,28),(9,730),(7,344)\}\end{aligned}$
$\therefore \quad$ Range of $f=\{0,1,28,730,344\}$
Find the values of $x$ for which the functions $f(x)=3 x^2-1$ and $g(x)=3+x$ are equal.
$\because \quad f(x)=g(x)$
$$\begin{aligned} & \Rightarrow \quad 3 x^2-1=3+x \\ & \Rightarrow \quad 3 x^2-x-4=0 \\ & \Rightarrow \quad 3 x^2-4 x+3 x-4=0 \\ & \Rightarrow \quad x(3 x-4)+1(3 x-4)=0 \\ & \Rightarrow \quad(3 x-4)(x+1)=0 \\ & \therefore \quad x=-1, \frac{4}{3} \end{aligned}$$
Is $g=\{(1,1),(2,3),(3,5),(4,7),\}$ a function, justify. If this is described by the relation, $g(x)=\alpha x+\beta$, then what values should be assigned to $\alpha$ and $\beta$ ?
We have, $$g=\{(1,1),(2,3),(3,5),(4,7)\}$$
Since, every element has unique image under $g$. So, $g$ is a function.
Now, $g(x)=\alpha x+\beta$
When $x=1$, then $g(1)=\alpha(1)+\beta\quad \text{.... (i)}$
$\Rightarrow \quad 1=\alpha+\beta$
When $x=2$, then $$g(2)=\alpha(2)+\beta$$
$$\Rightarrow \quad 3=2 \alpha+\beta\quad \text{.... (ii)}$$
On solving Eqs. (i) and (ii), we get
$$\alpha=2, \beta=-1$$