Inverse functions and derivatives

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agelessdrifter
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Inverse functions and derivatives

Postby agelessdrifter » Mon Oct 05, 2009 8:23 pm UTC

I feel silly for asking this question because I feel like I should know this by now (which is what I expect my calc1 professor to tell me when I approach her about it,) but I've looked through all of my old notes about inverse functions, and searched the internet for answers and haven't found any.

The question that's got me stuck is this: Find the value of (f-1)'(x) for the given function (f) and the real number, a.
f(x)=(1/27)(x5+2x3), a=11

I know that (f-1)'(x)= 1/(f'(f-1)(x)) and all that, but it suddenly occurred to me when I went to solve this problem that I don't know how to find the inverse of the given function. It's one-to-one, so it seems like it must have one. But I only know how to solve for f-1(x) when the given function only has instances of x which are all of the same degree (or can be algebraically manipulated to be all the same degree) or when f(x) can be worked out to be the power of some binomial quantity (like I can solve for the inverse of x2+2x+1 because it can be compressed to (x+1)2. Every tutorial or help site I can find online about inverse functions offers only help on these kinds of functions.

I also found it odd that the derivative of the given function, f(x)=(1/27)(x5+2x3) (which is (1/27)(5x4+6x2)), is not a 1-to-1 function, and should not, therefore, have an inverse. Is that part of the problem here? Thanks in advance for any help or advice.
Last edited by agelessdrifter on Mon Oct 05, 2009 8:39 pm UTC, edited 1 time in total.

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NathanielJ
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Re: Inverse functions and derivatives

Postby NathanielJ » Mon Oct 05, 2009 8:38 pm UTC

Don't try to find the inverse function for polynomials of order greater than 2 (the order in the case is 5) -- it's extremely messy if it even exists. Just note that for this question, you don't need to be able to find an expression for the inverse function, but rather you only need to know its value at x = 11. So solve:

11 = (1/27)(x5+2x3)

for x and go from there. Do you know how to solve this type of problem for x? It has a very nice solution in this case.
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HenryS
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Re: Inverse functions and derivatives

Postby HenryS » Mon Oct 05, 2009 8:43 pm UTC

I think you're supposed to use the result that (f-1)'(x) = (f'(x))-1. (The inverse function graph is a reflection of the original function graph along the line y=x, so you can tell what happens to the reflected tangent line as well: the slope of the new tangent is the inverse of the slope of the old.)

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agelessdrifter
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Re: Inverse functions and derivatives

Postby agelessdrifter » Mon Oct 05, 2009 9:53 pm UTC

NathanielJ wrote:Don't try to find the inverse function for polynomials of order greater than 2 (the order in the case is 5) -- it's extremely messy if it even exists. Just note that for this question, you don't need to be able to find an expression for the inverse function, but rather you only need to know its value at x = 11. So solve:

11 = (1/27)(x5+2x3)

for x and go from there. Do you know how to solve this type of problem for x? It has a very nice solution in this case.


Ughhh, that's right -- thanks so much. I was misunderstanding the theorem. I thought I needed to find f-1(11) and then (f-1)'(11) would be 1/(f'(f-1)(11)) (and I was wondering in what way that was supposed to be a useful theorem, :lol: )

Thanks again.


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