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Mathematics 18 Online
mercvriii:

What is the domain of the function f(x) = 2/3x – 7 when the range is {–3, 1, 5}?

umm:

Is there answer choices?

mercvriii:

@umm No.

KAPALOTJAY:

what do u think the answer is

mercvriii:

@KAPALOTJAY I don't know what the answer is. I need someone to explain it to me since I am perplexed.

KAPALOTJAY:

We find f(-3)=(-3)^2+3(-3)-7 = 9-9-7 =-7 then f(2)= 2^2+3(2)-7 = 4+6-7 =3 then f(5)= 5^2+3(5)-7 =25+15-7 = 33 the rang ={-7,3,33}

KAPALOTJAY:

range*

mercvriii:

@KAPALOTJAY Thank you.

KAPALOTJAY:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @mercvriii @KAPALOTJAY Thank you. \(\color{#0cbb34}{\text{End of Quote}}\) you are welcome

KAPALOTJAY:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @mercvriii @KAPALOTJAY Thank you. \(\color{#0cbb34}{\text{End of Quote}}\) best response?

KAPALOTJAY:

thx

mercvriii:

@KAPALOTJAY Sure!

umm:

\[f(-3)=(-3)^2+3(-3)-7\]\[= 9-9-7\]\[=-7\] \[f(2)= 2^2+3(2)-7\]\[= 4+6-7\]\[ =3\] \[f(5)= 5^2+3(5)-7\]\[=25+15-7\]\[= 33\] Therefore\[ ={-7,3,33}\]

umm:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @KAPALOTJAY We find f(-3)=(-3)^2+3(-3)-7 = 9-9-7 =-7 then f(2)= 2^2+3(2)-7 = 4+6-7 =3 then f(5)= 5^2+3(5)-7 =25+15-7 = 33 the rang ={-7,3,33} \(\color{#0cbb34}{\text{End of Quote}}\) Quote your sources or that's considered plagiarism--

darkknight:

umm, the problem asks What is the domain of the function f(x) = 2/3x – 7 when the range is {–3, 1, 5}? So you would plug in -3, 1 and 5 for f(x) and solve for x

mercvriii:

@umm @darkknight Thank you! :)

mercvriii:

@darkknight Thanks this helped!

darkknight:

np

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