anonymous
  • anonymous
if f(x) = x^2 -3x and g(x) = (√x)-1, then f(g(x)) =
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Nnesha
  • Nnesha
f(g(x)) can be written as f(sqrt{x}-1) so replace x with g(x) function into f(x) equation
anonymous
  • anonymous
i did but i still dont get it unfortunatey
Nnesha
  • Nnesha
\[\huge\rm f(x)=x^2-3x\] replace x with sqrt{X} -1

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Nnesha
  • Nnesha
what did you get ? can i see ur work plz
anonymous
  • anonymous
i got ((√x)-1)^2 - 3((√)-1)
Nnesha
  • Nnesha
\[\huge\rm (\sqrt{x}-1)^2-3(\sqrt{x}-1)\] distribute 2nd parentheses by -3 and apply foil method for (sqrt{x}-1)^2
Nnesha
  • Nnesha
do you know how to distribute ?-3(Sqrt{X}-1) ?
anonymous
  • anonymous
no i dont unfortunately. I dont know how to distribute the -3 to the sqaure roots or how to do the first set of parenthesis
Nnesha
  • Nnesha
apply distributive property distribute parentheses by outside number/ \[\huge\rm \color{reD}{a}(b+c) =\color{red}{a} ·b +\color{reD}{a}·c =\color{ReD}{a}b +\color{ReD}{a}c\]
Nnesha
  • Nnesha
\[-3(\sqrt{x}-1)\] multiply -3 times sqrt{x} and then -3 times -1
anonymous
  • anonymous
oh okay. I got this for the second set of parenthesis: -3√x+3
anonymous
  • anonymous
how do i foil square roots?
Nnesha
  • Nnesha
yes right now foil (sqrt{x} -1)^2 is same as \[\huge\rm (\sqrt{x}-1)(\sqrt{x}-1)\]|dw:1442527466021:dw| multiply 2nd parentheses by by first term sqrt{x} and then by -1
anonymous
  • anonymous
i ended up getting : √2x -√x -√x +1
anonymous
  • anonymous
i mean: √x^2 - √2x +1
Nnesha
  • Nnesha
\[\sqrt{2x}\]isn't right
Nnesha
  • Nnesha
|dw:1442528022410:dw| it would -2sqrt{x}
Nnesha
  • Nnesha
just like variables when we combine them we should coefficient add/subtract their coefficient
anonymous
  • anonymous
so it isnt √x^2 - √2x +1 ?
Nnesha
  • Nnesha
just like variables when we combine them we should coefficient add/subtract their coefficient so \[-\sqrt{x}-\sqrt{x}=-2\sqrt{x}\]
anonymous
  • anonymous
for the first set of parenthesis
Nnesha
  • Nnesha
2nd term isn't right
anonymous
  • anonymous
oh okay i see. so is it: √x^2 - 2√x +1
Nnesha
  • Nnesha
yes right now square root can cancel out the square so sqrt{x^2} = ??
anonymous
  • anonymous
okay so it would be x - √2x +1 -3√x +3???
Nnesha
  • Nnesha
yes right \[\huge\rm x-2\sqrt{x}+1-3\sqrt{x}+3\] now combine like terms
anonymous
  • anonymous
so it would be x-5√x+4?
Nnesha
  • Nnesha
yes looks right!
anonymous
  • anonymous
thank you very much for your help :)
Nnesha
  • Nnesha
yw :=)
Nnesha
  • Nnesha
did a great job!

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