marcelie
  • marcelie
help please !!!! find the inverse f(x) 3x^2+5 please check my work
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
marcelie
  • marcelie
|dw:1443237553803:dw|
steve816
  • steve816
that is some really messed up handwriting xD
marcelie
  • marcelie
the answer in my textbook says it -

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steve816
  • steve816
I wish I can help you, but it is too hard to read my friend.
marcelie
  • marcelie
sorry lol can you solve it ?
steve816
  • steve816
But to find the inverse, simply switch the x and y and solve for y.
marcelie
  • marcelie
k one sec let me rewrite it
steve816
  • steve816
subtract 5, divide by 3, square root it.
marcelie
  • marcelie
|dw:1443237871401:dw|
marcelie
  • marcelie
|dw:1443238092545:dw|
marcelie
  • marcelie
why does my textbook answer has the negative infront of the equation
steve816
  • steve816
Because you messed up a little bit.
marcelie
  • marcelie
oh
marcelie
  • marcelie
im confused ..
steve816
  • steve816
Don't be. Life has many confusing moments.
marcelie
  • marcelie
yeah
steve816
  • steve816
But never give up and keep trying until you are successful!
marcelie
  • marcelie
\[f (x ) = 3x ^{2}+5\]
marcelie
  • marcelie
\[x= 3y ^{2}+5\]
steve816
  • steve816
3y^2 = x-4
steve816
  • steve816
\[y = \frac{ \sqrt{x-5} }{ \sqrt{3} }\]
marcelie
  • marcelie
so do i first divide then square root it ? because i always get confused
steve816
  • steve816
You have to rationalize the denominator but I'm too lazy you can do that :)
marcelie
  • marcelie
isnt the answer that ?
steve816
  • steve816
no, you can't have a square root in the denominator.
marcelie
  • marcelie
mmk bc my solution has that answer and i dont understand why has negative
marcelie
  • marcelie
1 Attachment
marcelie
  • marcelie
so thats what i did
DanJS
  • DanJS
\[\sqrt{\frac{ a }{ b }}=\frac{ \sqrt{a} }{ \sqrt{b} }\]
DanJS
  • DanJS
recall, when you solve y = x^2 x = +root(y) or x= - root(y)
marcelie
  • marcelie
oh , so then i have to write -/+
marcelie
  • marcelie
soo every inverse has to have that -/+ when i take the sq root ?
DanJS
  • DanJS
no, that is just a property of square roots
DanJS
  • DanJS
but you have to figure which is the right answer
marcelie
  • marcelie
oh bc most of the problems ive done are sq roots but my solotuin book doesnt include -/+
marcelie
  • marcelie
how do i figure that out
DanJS
  • 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
  • DanJS
and a 3rd recall, the absolute value function is defined |dw:1443239752302:dw|
DanJS
  • DanJS
that is where the negative comes in
DanJS
  • 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
  • DanJS
notice the intervals they specified, they include all real numbers, but why break it up to that...
DanJS
  • 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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