Finding some critical points [basic algebra difficulty]

[K29]

Homework Statement

Had an optimization test today and had a problem with this question. I needed to find the stationary points of this function:
$$f(x)=\sqrt{x^{4}+1}-x+3$$

The Attempt at a Solution

So you differentiate that puppy with some chain rule for the square root part and tidy up you get:
$$f'(x)=4x^{6}-x^{4}-1$$
So to find some critical points you set that to 0... uh ok uhm now what?

 

[Mark44]

Homework Statement

Had an optimization test today and had a problem with this question. I needed to find the stationary points of this function:
$$f(x)=\sqrt{x^{4}+1}-x+3$$

The Attempt at a Solution

So you differentiate that puppy with some chain rule for the square root part and tidy up you get:
$$f'(x)=4x^{6}-x^{4}-1$$
So to find some critical points you set that to 0... uh ok uhm now what?

I think you "tidied" up way too much. After you have found the derivative, don't simplify it - just set it to zero and solve for x.

 

[Rasalhague]

You've got the right idea, but it looks to me like you may have gone astray with the simplifying. Once you're satisfied you've got the derivative, set it to 0, as you say, then solve for x. Why? Because that tells you what x is when the derivative is 0, which tells you what x is when your original function is stationary (not increasing or decreasing).

 

[K29]

Thanks guys.. but...
I'm still not able to pop out some decent x-values. After differentiating I have, untidied:
$$\frac{1}{2}4x^{3}(x^{4}+1)^{-\frac{1}{2}}-1=0$$

a bit neater:
$$\frac{2x^{3}}{\sqrt{x^{4}+1}}-1=0$$

Then:
$$2x^{3}-\sqrt{x^{4}+1}=0$$

I'm guessing one of these will probably be the best to work with. But I am still not seeing how to solve any of these forms for x :(
[edit: just noticed something donnt reply yet..lol]
[edit: nope, dead end..]

 

[lawtonfogle]

Take the (sorry, forgotten how to latex) 4x**6 - x**4 - 1 and substitute y = x**2.

So we then have a cubic formula.

Solve the cubic formula for what y must equal, then set x to be all of the possitive and negative square-roots of all give y's.

Wikipedia has the functions:
http://en.wikipedia.org/wiki/Cubic_function

let a = 4, b = -1, c = 0, and d = -1 and then plug into the following:

EDIT: Granted, there may be something I missed and this solution is totally bogus. I have been without a math class this entire semester (meh, that's what I get for making my second major Psychology and not Mathematics or Physics).

 

[Rasalhague]

Take the (sorry, forgotten how to latex) 4x**6 - x**4 - 1 and substitute y = x**2.

So we then have a cubic formula.

Solve the cubic formula for what y must equal, then set x to be all of the possitive and negative square-roots of all give y's.

I think the problem with starting out from 4x⁶ - x⁴ - 1 is that because this is derived by squaring both sides of the equation to get rid of the square root, it allows the possibility that what was squared could have had a negative value, whereas the square root function, by convention, only has positive values. (The actual inverse of squaring isn't a function, since it doesn't have a single, unambiguous output for each input.) Hence the difference between these graphs of the derivative, as calculated by Wolfram Alpha, and the "simplification"

http://www.wolframalpha.com/input/?i=2x^3-sqrt%28x^4%2B1%29
http://www.wolframalpha.com/input/?i=4x^6-x^4-1

The first one must be right, judging by its graph of the original function:

http://www.wolframalpha.com/input/?i=sqrt%28x^4%2B1%29-x%2B3

I must admit, I don't know how to solve for x in the derivative by hand; I just used a computer. Maybe someone more knowledgeable can help out there? I tried approximating it using the Taylor series expanded at 0, again letting Wolfram Alpha do the legwork. I got what I think is the right answer (to 4 decimal places), but also one other real solution, which can't be right, unless I've misunderstood something. Obviously I have a lot to learn...

 

[lawtonfogle]

While the functions themselves are different, would the area where they the x-axis be different (the second function crosses it twice, but we can throw away the negative case and only look a the positive case, as the negative case would give imaginary results anyways.

Wolfram says they are the same up to about 6 decimal places (it doesn't display any more than 6 decimal places).

Also, the solution given by my idea (substitute y = x**2, get the real points, take their square root to give you the points for x) also gives the same answer up to 6 decimal places (once again, it does not give more than that many).

The only problem is that you get the positive and negative, so there may be some work by hand to show the negative does not apply, but that is better than nothing, no?

 

[Rasalhague]

While the functions themselves are different, would the area where they the x-axis be different (the second function crosses it twice, but we can throw away the negative case and only look a the positive case, as the negative case would give imaginary results anyways.

Wolfram says they are the same up to about 6 decimal places (it doesn't display any more than 6 decimal places).

Also, the solution given by my idea (substitute y = x**2, get the real points, take their square root to give you the points for x) also gives the same answer up to 6 decimal places (once again, it does not give more than that many).

The only problem is that you get the positive and negative, so there may be some work by hand to show the negative does not apply, but that is better than nothing, no?

Yeah, if all we need is the stationary point, we can get the positive case from this and rule out the other by inspection. I guess the other real value given by the taylor series was a similar problem, although I don't really understand how that works.

If you type "taylor series ... to order n" you can get more terms. I went up to 39, and that was correct to 4 decimal places. I'm not sure how far you can go; with a few tens more than that, it'll give the approximation okay but it seems to be too complicated for it to solve for x.

 

[K29]

Thanks a lot guys. In the past few days I had a few other things to do..But now I went back onto the problem and sort of completed it with the Taylor Expansion. Not quite there but almost-I'm using the expansion in conjunction with some other stuff we learned in the course. The Taylor expansion is definitely the way to go. Even though its an approximation, it is perfect in our optimization course, because one of the intro chapters gives the proof for the expansion, so I assume that when we are confronted with a difficult equation like that to solve we should use that method.

Thanks again.