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