find inverse of y=(x+3)^2

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find inverse of y=(x+3)^2

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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@madhu.mukherjee.946
two equal signs ?
(x+3)^2 or (x-3)^2 ??

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(x+3)^2
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
@Nnesha pls help me with a stats question i'm stuck i've already tagged u
stat >.>
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
wait a sec i said something wrong
\[f(x)^{-1} =\sqrt{x}-3 \]for x>3 and \[f(x)^{-1} =-\sqrt{x}-3\] for x<3
how do yo get that answer @dinamix
only i use my mind cuz its easy for me this things
@Nnesha did u see my answer
No.
what are the steps to getting the answer
for ops -3 not 3 sorry i type cery fast
how he got the answer: take square root both sides
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
\[\sqrt{[x]^2} = |x|\]
did u know this rule right ?
yes ?
\[\sqrt{(x)^2} = |x|\] did u know this rule @Nnesha
yes. I said NO .cuz i didn't know how you got +3
\[y = (x+3)^2---> \sqrt{y}=|x+3| --> x=\sqrt{y}-3 --> f(x)^{-1} = \sqrt{x}-3\]
this for only when x>-3
did u see my steps ?
yes \[\huge\rm \sqrt{ x}=\pm \sqrt{(y+3)^2}\] just in case if he isn't familiar with the absolute value
so my method is wrong @Nnesha o.O ?
i didn't say that :=)
so what solution @Nnesha i think i am right cuz no mistakes
yes i think so. good job and thanks
u know how do it for x<-3 right ?@Nnesha
yes i do
@freckles my method is right ? look my answer plz
its not 3 i forget - okey
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
i didn't say tht ur answer is incorrect ....
\(\color{blue}{\text{Originally Posted by}}\) @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.
lol why do that @Nnesha

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