Question

Find the range if the function for each given domain f(x)=2x+3;[-2,-1,0,1,2]

Answer 1

Given:-

• 
  `\sf{f ( x ) = 2x + 3 - - - eqⁿ}`

`\:`

• 
  `\sf{Domains = \bold{[-2 , -1 , 0 , 1 , 2]}}`

`\:`

To find:-

• 
  `\sf{Range \: of \: the \: function = {?} }`

`\:`

`\underline{ \sf \: 1 ] \: f ( x ) = 2x + 3}`

now , put the value of x = -2 in eqⁿ

`\sf{↣f ( x ) = 2x + 3}`

`\sf{↣ f ( -2 ) = 2( -2 ) + 3}`

`\sf{↣ f ( - 2 ) = -4 + 3}`

`\boxed{ \sf \red{↣ f ( -2 ) = -1}}`

`\:`

━━━━━━━━━━━━━━━━━━━━━━━━━━━

`\underline{ \sf{ \: 2] \: f ( x ) = 2x + 3 \: }}`

put the value of x = -1 in eqⁿ

`\sf{↣f ( x ) = 2x + 3}`

`\sf{↣f ( -1 ) = 2( -1 ) + 3}`

`\sf{↣f ( -1 ) = -2 + 3}`

`\boxed{ \sf \color{green}↣f ( -1 ) = 1 }`

`\:`

━━━━━━━━━━━━━━━━━━━━━━━━━━━

`\underline{ \sf{ \: 3 ] \: f ( x ) = 2x + 3 \: }}`

put the value of x = 0 in eqⁿ

`\sf{↣f ( x ) = 2x + 3}`

`\sf{↣f ( 0 ) = 2( 0 ) + 3}`

`\sf{↣f ( 0 ) = 0 + 3}`

`\boxed{ \sf \blue{↣f ( 0 ) = 3}}`

`\:`

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

`\underline{ \sf{ \: 4 ] \: f ( x ) = 2x + 3 \: }}`

put the value of x = 1 in eqⁿ

`\sf{↣f ( x ) = 2x + 3}`

`\sf{↣f ( 1 ) = 2( 1 ) + 3}`

`\sf{↣f ( 1 ) = 2 + 3}`

`\boxed { \sf \color{brown}↣f ( 1 ) = 5}`

`\:`

━━━━━━━━━━━━━━━━━━━━━━━━━━━

`\underline {\sf{ \:5 ] \: f ( x ) = 2x + 3 \: }}`

put the value of x = 2 in eqⁿ

`\sf{↣f ( x ) = 2x + 3}`

`\sf{↣f ( 2 ) = 2( 2 ) + 3}`

`\sf{↣f ( 2 ) = 4 + 3}`

`\boxed{ \sf \purple{↣f ( 2 ) = 7}}`

`\:`

━━━━━━━━━━━━━━━━━━━━━━━━━━

Range of the function for each domain is:-

• 
  `\bold{ \underline{ \boxed{ \sf \blue{-1, 1, 3, 5, 7.}}}}`

`\:`

hope it helps! :)