Question

Helppp

The functions f and g are defined as follows:

Answer 1

#a

`\\ \rm\Rrightarrow f(-4)`

`\\ \rm\Rrightarrow 3(-4)+2`

`\\ \rm\Rrightarrow -12+2`

`\\ \rm\Rrightarrow -10`

#b

#1

`\\ \rm\Rrightarrow y=(2x-1)/(3)`

• Interchange x,y

`\\ \rm\Rrightarrow x=(2y-1)/(3)`

Find y

`\\ \rm\Rrightarrow y=(3x+1)/(2)`

Inverse is

`\\ \rm\Rrightarrow g^(-1)(x)=(3x+1)/(2)`

#2

`\\ \rm\Rrightarrow gof(x)`

`\\ \rm\Rrightarrow g(f(x))`

`\\ \rm\Rrightarrow g(3x+2)`

`\\ \rm\Rrightarrow (2(3x+2)-1)/(3)`

`\\ \rm\Rrightarrow (6x+4-1)/(3)`

`\\ \rm\Rrightarrow (6x+3)/(3)`

If we factor out

`\\ \rm\Rrightarrow (2x+1)/(1)`

`\\ \rm\Rrightarrow 2x+1`

#c

`\\ \rm\Rrightarrow f(x)=g(x)`

`\\ \rm\Rrightarrow 3x+2=(2x-1)/(3)`

`\\ \rm\Rrightarrow 3(3x+2)=2x-1`

`\\ \rm\Rrightarrow 9x+6=2x-1`

`\\ \rm\Rrightarrow 7x=-7`

`\\ \rm\Rrightarrow x=-1`

Answer 2

Given functions:

`f(x)=3x+2`

`g(x)=\left((2x-1)/(3)\right)`

Part (a)

`\begin{aligned}\implies f(-4) & = 3(-4)+2\\& = -12+2\\ & = -10\end{aligned}`

Part (b)(i)

`\begin{aligned}g(x) & =\left((2x-1)/(3)\right)\\\\\textsf{Swap }g(x) \textsf{ for }y : \\\implies y & = \left((2x-1)/(3)\right)\\\\\textsf{Make } x \textsf{ the subject}: \\\implies 3y & = 2x-1\\3y+1 & = 2x\\x & = (3y+1)/(2)\\\\\textsf{Swap }x \textsf{ for }g^(-1)(x) \textsf{ and }y \textsf{ for }x:\\\implies g^(-1)(x) & = (3x+1)/(2)\end{aligned}`

Part (b)(ii)

`\begin{aligned}gf(x) & = (2[f(x)]-1)/(3)\\\\& = (2(3x+2)-1)/(3)\\\\& = (6x+4-1)/(3)\\\\& = (6x+3)/(3)\\\\& = (6x)/(3)+(3)/(3)\\\\& = 2x+1\end{aligned}`

Part (c)

`\begin{aligned}f(x) & = g(x)\\\\\implies 3x+2 & = (2x-1)/(3)\\\\3(3x+2) & = 2x-1\\\\9x+6 & = 2x-1\\\\7x & = -7\\\\\implies x & = -1\end{aligned}`