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

Hi
I need to find the inverse of the function $$f\left( x \right) = x^2 + 6x\$$ with the domain of $$x \ge - 3$$ and $$x \le - 3$$.

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

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

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AlephZero
Homework Helper
$$y = x^2 + 6x$$
and then you want to get equation x = some function of y?

$$x^2 + 6x - y = 0$$ is just a quadratic equation in x.

Gib Z
Homework Helper
No No you don't just swap the x and y in one case, it must be all the cases.

so the $$f(x)=y=x^2 + 6x$$ Then $$x=y^2 + 6y$$ is your inverse.

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

dextercioby
Homework Helper
so the $$f(x)=y=x^2 + 6x$$ Then $$x=y^2 + 6y$$ is your inverse.

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

Gib Z
Homework Helper
>.< Well i thought what i said was correct, but if dexter says it isn't, then im wrong for sure :). BUt yea im 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.

HallsofIvy
Homework Helper
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 ____________.

Gib Z
Homework Helper
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)

dextercioby
Homework Helper
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.

Gib Z
Homework Helper
ahh ok thanks for telling me that :) Didnt realise x could be treated like that. ty

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 $$x \ge - 3\,\,\,\,\,and\,\,\,\,\,x \le - 3$$ will be:

$$f^{ - 1} \left( x \right) = - 3 \pm \sqrt {36 - 4x}$$

?
many thanks

Gib Z
Homework Helper

$$y(x)=\frac{-6 \pm \sqrt {36-4x}}{2}$$, which is definitely not what you have.

cristo
Staff Emeritus
the domain of $$x \ge - 3\,\,\,\,\,and\,\,\,\,\,x \le - 3$$ will be:
What does this domain mean? Is there a typo here; as it stands, this is equal to R

dextercioby
Homework Helper
I get

$$y(x)=-3\pm \sqrt{9+x}$$

ah i see where i have gone wrong.
thanks for the help, and sorry for any inconvenience

Gib Z
Homework Helper
I get

$$y(x)=-3\pm \sqrt{9+x}$$
Well yea, thats after simplifications >.< I was still correct, technically :P.

dextercioby