Question

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

Answer 1

@madhu.mukherjee.946

Answer 2

two equal signs ?

Answer 3

(x+3)^2 or (x-3)^2 ??

Answer 4

(x+3)^2

Answer 5

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

Answer 6

@Nnesha pls help me with a stats question i'm stuck i've already tagged u

Answer 7

stat >.>

Answer 8

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

Answer 9

wait a sec i said something wrong

Answer 10

\[f(x)^{-1} =\sqrt{x}-3 \]for x>3
and \[f(x)^{-1} =-\sqrt{x}-3\] for x<3

Answer 11

how do yo get that answer @dinamix

Answer 12

@sarahm22

Answer 13

only i use my mind cuz its easy for me this things

Answer 14

@Nnesha did u see my answer

Answer 15

No.

Answer 16

what are the steps to getting the answer

Answer 17

for ops -3 not 3 sorry i type cery fast

Answer 18

how he got the answer:
take square root both sides

Answer 19

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

Answer 20

\[\sqrt{[x]^2} = |x|\]

Answer 21

did u know this rule right ?

Answer 22

@Nnesha

Answer 23

yes ?

Answer 24

\[\sqrt{(x)^2} = |x|\] did u know this rule @Nnesha

Answer 25

yes.
I said NO .cuz i didn't know how you got +3

Answer 26

\[y = (x+3)^2---> \sqrt{y}=|x+3| --> x=\sqrt{y}-3 --> f(x)^{-1} = \sqrt{x}-3\]

Answer 27

@Nnesha

Answer 28

this for only when x>-3

Answer 29

did u see my steps ?

Answer 30

yes \[\huge\rm \sqrt{ x}=\pm \sqrt{(y+3)^2}\]
just in case if he isn't familiar with the absolute value

Answer 31

so my method is wrong @Nnesha o.O ?

Answer 32

i didn't say that :=)

Answer 33

so what solution @Nnesha i think i am right cuz no mistakes

Answer 34

yes i think so.
good job and thanks

Answer 35

u know how do it for x<-3 right ?@Nnesha

Answer 36

@Nnesha

Answer 37

yes i do

Answer 38

@freckles my method is right ? look my answer plz

Answer 39

its not 3 i forget - okey

Answer 40

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

Answer 41

i didn't say tht ur answer is incorrect ....

Answer 42

\(\color{blue}{\text{Originally Posted by}}\) @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.

Answer 43

lol why do that @Nnesha