anonymous
  • anonymous
find inverse of y=(x+3)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@madhu.mukherjee.946
Nnesha
  • Nnesha
two equal signs ?
Nnesha
  • Nnesha
(x+3)^2 or (x-3)^2 ??

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More answers

anonymous
  • anonymous
(x+3)^2
Nnesha
  • Nnesha
okay cool same like last one (x+3)^2 is same as (x+3)(x+3) so apply the foil method to expand (x+3)^2 and then replace x with y and y with x
madhu.mukherjee.946
  • madhu.mukherjee.946
@Nnesha pls help me with a stats question i'm stuck i've already tagged u
Nnesha
  • Nnesha
stat >.>
anonymous
  • anonymous
so i foiled (x+3)^2 I got x^2+6x+9 so if you switch the x and y you would get x=y^2+6y-9
Nnesha
  • Nnesha
wait a sec i said something wrong
dinamix
  • dinamix
\[f(x)^{-1} =\sqrt{x}-3 \]for x>3 and \[f(x)^{-1} =-\sqrt{x}-3\] for x<3
anonymous
  • anonymous
how do yo get that answer @dinamix
dinamix
  • dinamix
@sarahm22
dinamix
  • dinamix
only i use my mind cuz its easy for me this things
dinamix
  • dinamix
@Nnesha did u see my answer
Nnesha
  • Nnesha
No.
anonymous
  • anonymous
what are the steps to getting the answer
dinamix
  • dinamix
for ops -3 not 3 sorry i type cery fast
Nnesha
  • Nnesha
how he got the answer: take square root both sides
Nnesha
  • Nnesha
that's what he did: replace x with y and y with x 2nd) then take square root both sides to cancel out the square
dinamix
  • dinamix
\[\sqrt{[x]^2} = |x|\]
dinamix
  • dinamix
did u know this rule right ?
dinamix
  • dinamix
@Nnesha
Nnesha
  • Nnesha
yes ?
dinamix
  • dinamix
\[\sqrt{(x)^2} = |x|\] did u know this rule @Nnesha
Nnesha
  • Nnesha
yes. I said NO .cuz i didn't know how you got +3
dinamix
  • dinamix
\[y = (x+3)^2---> \sqrt{y}=|x+3| --> x=\sqrt{y}-3 --> f(x)^{-1} = \sqrt{x}-3\]
dinamix
  • dinamix
@Nnesha
dinamix
  • dinamix
this for only when x>-3
dinamix
  • dinamix
did u see my steps ?
Nnesha
  • Nnesha
yes \[\huge\rm \sqrt{ x}=\pm \sqrt{(y+3)^2}\] just in case if he isn't familiar with the absolute value
dinamix
  • dinamix
so my method is wrong @Nnesha o.O ?
Nnesha
  • Nnesha
i didn't say that :=)
dinamix
  • dinamix
so what solution @Nnesha i think i am right cuz no mistakes
Nnesha
  • Nnesha
yes i think so. good job and thanks
dinamix
  • dinamix
u know how do it for x<-3 right ?@Nnesha
dinamix
  • dinamix
@Nnesha
Nnesha
  • Nnesha
yes i do
dinamix
  • dinamix
@freckles my method is right ? look my answer plz
dinamix
  • dinamix
its not 3 i forget - okey
freckles
  • freckles
I guess he was looking for inverse relation since an inverse function doesn't exist. And yeah... \[y=(x+3)^2 \\ \text{ take square root of both sides } \\ \pm \sqrt{y}=x+3 \\ \text{ now if } x+3 \text{ is positive or zero } (x+3 \ge 0) \\ \text{ then } x+3=\sqrt{y} \\ \text{ and if } x+3 \text{ is negative } \\ \text{ then } x+3=-\sqrt{y} \] \[\text{ solving each } \\ x=\sqrt{y}-3 \text{ if } x+3 \ge 0 \text{ or solving the inequality we get } x \ge -3 \\ x=-\sqrt{y}-3 \text{ if } x+3 <0 \text{ or solving that inequality we get } x<-3 \\ \text{ so we have } y=\sqrt{x}-3 \text{ if } x \ge -3 \\ \text{ or } y=-\sqrt{x}-3 \text{ if } x<-3 \] which you already said
Nnesha
  • Nnesha
i didn't say tht ur answer is incorrect ....
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @dinamix its not 3 i forget - okey \(\color{blue}{\text{End of Quote}}\) \(\color{blue}{\text{Originally Posted by}}\) @dinamix for ops -3 not 3 sorry i type cery fast \(\color{blue}{\text{End of Quote}}\) i know.
dinamix
  • dinamix
lol why do that @Nnesha

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