help please !!!! find the inverse
f(x) 3x^2+5
please check my work

- marcelie

help please !!!! find the inverse
f(x) 3x^2+5
please check my work

- schrodinger

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- marcelie

|dw:1443237553803:dw|

- steve816

that is some really messed up handwriting xD

- marcelie

the answer in my textbook says it -

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## More answers

- steve816

I wish I can help you, but it is too hard to read my friend.

- marcelie

sorry lol can you solve it ?

- steve816

But to find the inverse, simply switch the x and y and solve for y.

- marcelie

k one sec let me rewrite it

- steve816

subtract 5, divide by 3, square root it.

- marcelie

|dw:1443237871401:dw|

- marcelie

|dw:1443238092545:dw|

- marcelie

why does my textbook answer has the negative infront of the equation

- steve816

Because you messed up a little bit.

- marcelie

oh

- marcelie

im confused ..

- steve816

Don't be. Life has many confusing moments.

- marcelie

yeah

- steve816

But never give up and keep trying until you are successful!

- marcelie

\[f (x ) = 3x ^{2}+5\]

- marcelie

\[x= 3y ^{2}+5\]

- steve816

3y^2 = x-4

- steve816

\[y = \frac{ \sqrt{x-5} }{ \sqrt{3} }\]

- marcelie

so do i first divide then square root it ? because i always get confused

- steve816

You have to rationalize the denominator but I'm too lazy you can do that :)

- marcelie

isnt the answer that ?

- steve816

no, you can't have a square root in the denominator.

- marcelie

mmk bc my solution has that answer and i dont understand why has negative

- marcelie

##### 1 Attachment

- marcelie

so thats what i did

- DanJS

\[\sqrt{\frac{ a }{ b }}=\frac{ \sqrt{a} }{ \sqrt{b} }\]

- DanJS

recall, when you solve y = x^2
x = +root(y) or x= - root(y)

- marcelie

oh , so then i have to write -/+

- marcelie

soo every inverse has to have that -/+ when i take the sq root ?

- DanJS

no, that is just a property of square roots

- DanJS

but you have to figure which is the right answer

- marcelie

oh bc most of the problems ive done are sq roots but my solotuin book doesnt include -/+

- marcelie

how do i figure that out

- DanJS

also another thing to note,
\[\sqrt{y^2} = \left| y \right|\]
the square root of y^2 will always be positive, and is really the absolute value of y, not just y

- DanJS

and a 3rd recall, the absolute value function is defined
|dw:1443239752302:dw|

- DanJS

that is where the negative comes in

- DanJS

Also as an aside, since you are (x,y) --->(y,x) swapping y and x,
The inverse, if there is one, is a reflection over the line y = x ; through the origin slope 1

- DanJS

notice the intervals they specified, they include all real numbers, but why break it up to that...

- DanJS

Do they give just the negative root as an answer or both + and - roots depending on which interval you are talking about in the domain

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