Zero derivative on a strictly increasing function (not piecewise)

[quincyboy7]

Homework Statement

Find a function, strictly increasing, whose derivative equals zero at TWO places. Find another such function whose derivative equals zero at infinitely many places. These should not be piecewise and should be continuous on all reals.

Homework Equations

f(x)=x^3 has one such spot, at x=0.

The Attempt at a Solution

I can't imagine the answer is a polynomial, since the derivatives of those will have x terms that can equal 0 at only one place (think x^3, x^5, x^7, etc.). Any manipulations of such polynomials make them not strictly increasing, which is another problem.

Sine and cose functions probably can't be made strictly increasing by any manipulations. Inverse functions don't have derivatives equaling zero, I don't think.

E^x and sqrt(x) are both not continuous on both reals and don't have derivative zero anywhere.

I just can't think of anything more in terms of function "families" that could work...any pointers would be great

 

[tiny-tim]

hi quincyboy7!

… I can't imagine the answer is a polynomial, since the derivatives of those will have x terms that can equal 0 at only one place (think x^3, x^5, x^7, etc.). Any manipulations of such polynomials make them not strictly increasing, which is another problem.

that's right

Sine and cose functions probably can't be made strictly increasing by any manipulations.

how about stretching one of them out by adding another function to it?

 

[quincyboy7]

how about stretching one of them out by adding another function to it?

Well, the function would still decrease even if it wasn't periodic, like cosx+x, right? There's no way sinx + FUNCTION doesn't go down at some interval, right? Even if that interval isn't regular...

EDIT: OH! The derivative of cosx+x is -sinx+1, which is 0 at infinitely many places and never negative! Thanks SO much.

But how would I get one that's "controllable" i.e. f'(x)=0 at only TWO places?

 

[tiny-tim]

min [d/dx (sinx + f(x))] = … ?

 

[henry_m]

Think about what the graph of the derivative of the function would look like.

BTW you can find an increasing polynomial with zero derivative at any finite number of points; it must be odd, and to have n such points must be of degree 2n+1 or higher. (You see why?). So for your first question, you need a quintic at least.

 

[quincyboy7]

Think about what the graph of the derivative of the function would look like.

BTW you can find an increasing polynomial with zero derivative at any finite number of points; it must be odd, and to have n such points must be of degree 2n+1 or higher. (You see why?). So for your first question, you need a quintic at least.

I understand why it must be odd, but I don't quite get how you got the 2n+1 expression, even though it seems intuitive.

Alright, let me try to integrate a function that's positive everywhere and zero at 2 places...like x^2(x-1)^2, which will be INT (x^4-2x^3+x^2) or x^5/5 -x^4/2 + x^3/3...does that work?

EDIT: Here's how I think you got the 2n+1...the derivative equals zero at n places, but never crosses the x-axis so the form of the derivative is (x-n1)^2 (at least, could be 4, 6, 8, etc.) (x-n2)^2 etc. The order of the derivative will thus be at least 2n and the integral will be 2n+1 at least.

 

[henry_m]

I understand why it must be odd, but I don't quite get how you got the 2n+1 expression, even though it seems intuitive.

Alright, let me try to integrate a function that's positive everywhere and zero at 2 places...like x^2(x-1)^2, which will be INT (x^4-2x^3+x^2) or x^5/5 -x^4/2 + x^3/3...does that work?

EDIT: Here's how I think you got the 2n+1...the derivative equals zero at n places, but never crosses the x-axis so the form of the derivative is (x-n1)^2 (at least, could be 4, 6, 8, etc.) (x-n2)^2 etc. The order of the derivative will thus be at least 2n and the integral will be 2n+1 at least.

Exactly!

In fact, by thinking about the derivatives like this, you can get even nastier things. For example, you can have a function which is strictly increasing, differentiable everywhere and has infinitely many stationary points in a finite interval:
$$f(x)=\int_0^x \sin(1/t) dt +x$$
[The tricky thing here is showing it's differentiable at the origin.]

 

[SammyS]

...
I can't imagine the answer is a polynomial, since the derivatives of those will have x terms that can equal 0 at only one place (think x^3, x^5, x^7, etc.). Any manipulations of such polynomials make them not strictly increasing, which is another problem.
...

If f(x) = x³, then f'(x) = 3x².

Look at g'(x) = (x-1)²(x+1)², which is zero at x = ±1, and positive otherwise.

Find g(x) by integration. Of course, g'(x) may be written as g'(x) = (x²-1)² = x⁴ -2x² +1 .