I need help with finding the inverse of this function, 2x^5+x^3+1.

- anonymous

I need help with finding the inverse of this function, 2x^5+x^3+1.

- katieb

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- anonymous

\[2x^5+x^3+1\]

- jdoe0001

hmm, technically, the inverse is just swapping about the variables, now the simplification, well, not so simplistic

- anonymous

exactly, I know the concept it's just that I have to show my work to simplify and insert a value into the new function.

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## More answers

- klimenkov

Interesting...
Can somebody find the inverse function for \(y=x^2+2x-3\) ?

- anonymous

yes... but you will have to restrict the domain.

- klimenkov

Two branches will satisfy:
\[-1\pm\sqrt{4+x}.\]Now how to solve this one?
\[2y^5+y^3+1-x=0.\]

- anonymous

throwing it into wolframalpha gives this result
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ef49hilhcgk

- klimenkov

This means that the inverse function may not be found.

- anonymous

whaaaaaat...... But It's in my textbook and part of my homework! Ugh...

- anonymous

what are you studying?

- anonymous

calc one... this is the second day of homework.

- anonymous

you sure it's a 5th power?

- anonymous

sorry, that's the class. I'm studying Electrical Engineering. just a second and I'll upload a snapshot of the problem

- anonymous

#74

##### 1 Attachment

- klimenkov

Hahaha. Lol.

- jdoe0001

hmm well well

- anonymous

dude... it says f^-1(a) and gives you the value of a.
inverse functions work like this...
plug in the y from a function into its inverse and you get out x.
compute f(a) for your functions and then f^-1(f(a)) will just be a.

- anonymous

I know, and I have the -71 as the answer, but it also asks to write down the function which I can't find to save my life!

- anonymous

oops... i misread. but i doesn't ask for the inverse function... only f^-1(a)

- anonymous

OH! oh oh oh oh.... Damn....

- anonymous

if you find a = f(x) then the x that satisfies this will be f^-1(a)

- DebbieG

Wow that's a lot easier than finding \(f^{-1}(x)\) :)

- anonymous

yeah, it's -71...

- anonymous

craaaaap.... well there went an hour I'll never get back. thanks for all your help guys

- anonymous

no worries... maybe post the question from your book and your interpretation so you can get to the heart of the sunrise quicker.

- klimenkov

So, what is an answer?

- anonymous

it's not -71

- anonymous

no?

- anonymous

x = -1
-2 = 2x^5 + x^3 +1 => 2x^5 + x^3 +3 = 0
the only real solution is x = -1

- anonymous

yeah I get what you mean now... but isn't that just f(-2) now? not the inverse?

- anonymous

no... f(-1) = -2 so
f^-1(-2) = -1

- DebbieG

Remember that \(f^{-1}(-2)\) is that domain element x that gives you \(f(x)=-2\)
The function maps the domain to the range.
The inverse function maps the range back to the domain.

- klimenkov

Let \(f^{-1}(a)=x\). Now map \(f\) to both of the parts of the equation and use \(f(f^{-1}(a))=a\). It will become\[a=f(x)\]Solve this and you will obtain an answer.

- DebbieG

So by setting \(f(x)=y=-2\) and solving, you find the x in the domain that takes you to y in the range.

- anonymous

f^-1(f(x)) = x

- anonymous

ohhh okay, so put the variable as the solution and work backwards

- anonymous

they give you a = f(x) and want you to find f^-1(a) = f^-1(f(x)) = x
so setting f(x) = a and solving for x gives you what you need.

- anonymous

make sense?

- anonymous

Yep :) thanks again y'all

- anonymous

you're welcome!

- anonymous

*lightbulb turns on* I think the rest of this will be much easier now

- anonymous

awesome! let's turn the dark into light!

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