Inverse of f(x)=x^2+6x

  • #1
unique_pavadrin
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Hi
I need to find the inverse of the function [tex]f\left( x \right) = x^2 + 6x\[/tex] with the domain of [tex]x \ge - 3[/tex] and [tex]x \le - 3[/tex].

What i have done so far:
[tex]
\begin{array}{c}
f\left( x \right) = x^2 + 6x \\
x = y^2 + 6x \\
\end{array}
[/tex]

and then from here I need to put the 2nd function in terms of y, however I do not know how to because of the function having the variable y twice. I am not asking for the answer of the problem, only suggestions on which way to approach the situation.

Many thanks
unique_pavadrin
 

Answers and Replies

  • #2
AlephZero
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Shouldn't your last equation be
[tex]y = x^2 + 6x[/tex]
and then you want to get equation x = some function of y?

[tex]x^2 + 6x - y = 0[/tex] is just a quadratic equation in x.
 
  • #3
Gib Z
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No No you don't just swap the x and y in one case, it must be all the cases.

so the [tex]f(x)=y=x^2 + 6x[/tex] Then [tex]x=y^2 + 6y[/tex] is your inverse.

I can't see any obvious ways to get y on its own from there.
 
  • #4
dextercioby
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so the [tex]f(x)=y=x^2 + 6x[/tex] Then [tex]x=y^2 + 6y[/tex] is your inverse.

I can't see any obvious ways to get y on its own from there.

?
 
  • #5
Gib Z
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>.< Well i thought what i said was correct, but if dexter says it isn't, then I am wrong for sure :). BUt yea I am pretty sure when you want the inverse, you change ALL the x's to ys and vice versa. in the OP the last bit of TEX was incorrect.
 
  • #6
HallsofIvy
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No, Gib Z, dextercioby's ? wasn't directed toward your saying that to find the inverse of y= x2+ 6x you start by changing it to x=y2+ 6y, it was toward "I can't see any obvious ways to get y on its own from there."

That's a quadratic function of y. You can always solve it by using the ____________.
 
  • #7
Gib Z
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Ahh I don't understand how the quadratic forumula would work >.<" Unless you mean take x to the other side and pretend its a constant? ...x=y(y+6)
 
  • #8
dextercioby
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Exactly. y^2 +6y-x=0 is a quadratic and should be solved for y(x) which is the inverse of the initial function, once one takes care of the domain issues.
 
  • #9
Gib Z
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ahh ok thanks for telling me that :) Didnt realize x could be treated like that. ty
 
  • #10
unique_pavadrin
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Thanks for all the post people, your help is greatly appreciated. Sorry, wasn't able to get online until now due to isp problems. In my 1st post, the 2nd equation was a typo, it should have read x=y^2+6y.

So therefore the inverse of the function between the domain of [tex]x \ge - 3\,\,\,\,\,and\,\,\,\,\,x \le - 3[/tex] will be:

[tex]f^{ - 1} \left( x \right) = - 3 \pm \sqrt {36 - 4x}[/tex]

?
many thanks
 
  • #11
Gib Z
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The quadratic equation yields:

[tex]y(x)=\frac{-6 \pm \sqrt {36-4x}}{2}[/tex], which is definitely not what you have.
 
  • #12
cristo
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the domain of [tex]x \ge - 3\,\,\,\,\,and\,\,\,\,\,x \le - 3[/tex] will be:

What does this domain mean? Is there a typo here; as it stands, this is equal to R
 
  • #13
dextercioby
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I get

[tex] y(x)=-3\pm \sqrt{9+x} [/tex]
 
  • #14
unique_pavadrin
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ah i see where i have gone wrong.
thanks for the help, and sorry for any inconvenience
unique_pavadrin
 
  • #15
Gib Z
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I get

[tex] y(x)=-3\pm \sqrt{9+x} [/tex]

Well yea, that's after simplifications >.< I was still correct, technically :P.
 
  • #16
dextercioby
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Up to the minus sign under the square root, yeah...
 
  • #17
Gib Z
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>.< O jesus Christ shoot me
 

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