Inverse problem that should be easy

[trajan22]

Homework Statement

f(x)=x^5-x^3+2x,
find inverse f(x)

]3. The Attempt at a Solution [/b]
ive not really gone anywhere on the problem all I've done is set y=x(x^4-x^2+2)
i can't really see any way to set x=y in this equation, what is the trick that I am simply not seeing?

 

[HallsofIvy]

My question is why you think this problem should be easy!

Basically, you are asked to solve the equation y= x⁵- x³+ 2x for any value of y. I see no reason to think that can be done.

 

[trajan22]

i thought there was some little trick that are usually associated with these kinds of problems. its in my book so i know that there must be some method of finding the inverse.

eventually I am supposed to use inverse g'(a)=1/(f'(g(a)) where g(a) = inverse of f (x) when a=2

i know this part above can be done without actually finding f(x)'s inverse directly but later on in the problem I am supposed to compare. which would require finding f(x)'s inverse

 

[matt grime]

You can't write down the inverse of f in any explicit fashion. For starters, it isn't even an invertible function since it is not one to one. Locally it is invertible, though, but you won't be able to write out the inverse in any nice representation.

 

[HallsofIvy]

matt, I graphed it and it looked like it was 1-1 to me! Can you give two values of x that have the same f value?

 

[matt grime]

Sorry, just presumed it would have some local maxima/minima. OK, in general a function like that is not invertible at all - but is locally.

Why graph it, though? What are the roots of 5x^4-3x^2+2?

 

[trajan22]

well 5x^4-3x^2+2 has no roots because it has a minimum around 2 (not quite 2). but other than this telling us what f is doing why take the derivitive? it shows that f has no critical points to take into consideration.

so how else would i go about solving this problem if i can't find an inverse and it says
calculate g(x) and state the domain and range
remember:(g(x)=inverse of f(x)

 

[trajan22]

sorry guys nevermind i looked through the book again to triple check that i was reading the directions and I am supposed to find the inverse of f '(a) so to find this would i take the derivitive of f and then find an inverse for f '(a) however my problem is that the derivitive gives a function that is not one to one. which basically takes me back to beginning

 

[matt grime]

well 5x^4-3x^2+2 has no roots because it has a minimum around 2 (not quite 2). but other than this telling us what f is doing why take the derivitive? it shows that f has no critical points to take into consideration.

Becuase it tells you that the function f is one to one (it was clearly onto already), and thus globally invertible.

Just look at the theorem you're proving - if the derivative vanished anywhere then yuou can't invert it with the theorem. More concretely just look at y=x^2. This is invertible on the positive real x, but not all the real line. Why? Becuase the derivative vanishes at 0. If you must, think about the so-called horizontal line test that I'm sure you've been told about.

 

[matt grime]

sorry guys nevermind i looked through the book again to triple check that i was reading the directions and I am supposed to find the inverse of f '(a) so to find this would i take the derivitive of f and then find an inverse for f '(a) however my problem is that the derivitive gives a function that is not one to one. which basically takes me back to beginning

You can still use the inverse function theoremm locally. Just look above at the post I gave for y=x^2. You're happy thinking of that as inveertible, but it isn't. Restrict the domain and it is invertible. Derivatives are a purely local object. Global behaviour doesn't matter when you want to verify something locally.

 

[trajan22]

right...so ill restrict the domain and take the inverse, I guess I am having an algebraic problem because i get y-2=5x^4-3x^2 from here I am having trouble finding a way to solve for x.

 

[slearch]

use the quadratic formula