anonymous
  • anonymous
I need help with finding the inverse of this function, 2x^5+x^3+1.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[2x^5+x^3+1\]
jdoe0001
  • jdoe0001
hmm, technically, the inverse is just swapping about the variables, now the simplification, well, not so simplistic
anonymous
  • 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
  • klimenkov
Interesting... Can somebody find the inverse function for \(y=x^2+2x-3\) ?
anonymous
  • anonymous
yes... but you will have to restrict the domain.
klimenkov
  • klimenkov
Two branches will satisfy: \[-1\pm\sqrt{4+x}.\]Now how to solve this one? \[2y^5+y^3+1-x=0.\]
anonymous
  • anonymous
throwing it into wolframalpha gives this result http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ef49hilhcgk
klimenkov
  • klimenkov
This means that the inverse function may not be found.
anonymous
  • anonymous
whaaaaaat...... But It's in my textbook and part of my homework! Ugh...
anonymous
  • anonymous
what are you studying?
anonymous
  • anonymous
calc one... this is the second day of homework.
anonymous
  • anonymous
you sure it's a 5th power?
anonymous
  • anonymous
sorry, that's the class. I'm studying Electrical Engineering. just a second and I'll upload a snapshot of the problem
anonymous
  • anonymous
klimenkov
  • klimenkov
Hahaha. Lol.
jdoe0001
  • jdoe0001
hmm well well
anonymous
  • 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
  • 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
  • anonymous
oops... i misread. but i doesn't ask for the inverse function... only f^-1(a)
anonymous
  • anonymous
OH! oh oh oh oh.... Damn....
anonymous
  • anonymous
if you find a = f(x) then the x that satisfies this will be f^-1(a)
DebbieG
  • DebbieG
Wow that's a lot easier than finding \(f^{-1}(x)\) :)
anonymous
  • anonymous
yeah, it's -71...
anonymous
  • anonymous
craaaaap.... well there went an hour I'll never get back. thanks for all your help guys
anonymous
  • 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
  • klimenkov
So, what is an answer?
anonymous
  • anonymous
it's not -71
anonymous
  • anonymous
no?
anonymous
  • anonymous
x = -1 -2 = 2x^5 + x^3 +1 => 2x^5 + x^3 +3 = 0 the only real solution is x = -1
anonymous
  • anonymous
yeah I get what you mean now... but isn't that just f(-2) now? not the inverse?
anonymous
  • anonymous
no... f(-1) = -2 so f^-1(-2) = -1
DebbieG
  • 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
  • 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
  • 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
  • anonymous
f^-1(f(x)) = x
anonymous
  • anonymous
ohhh okay, so put the variable as the solution and work backwards
anonymous
  • 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
  • anonymous
make sense?
anonymous
  • anonymous
Yep :) thanks again y'all
anonymous
  • anonymous
you're welcome!
anonymous
  • anonymous
*lightbulb turns on* I think the rest of this will be much easier now
anonymous
  • anonymous
awesome! let's turn the dark into light!

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