# Analysis of a function(relatively easy)

1. Oct 15, 2009

### Wingeer

1. The problem statement, all variables and given/known data

Given the function:
$$f(x)=\left\{ \begin{array}{cl} x+2x^2\sin\left(\frac{1}{x}\right), & x\neq 0 \\ 0, & x=0 \end{array}\right.$$
& $$f'(0)=1$$

"Show that all intervals containing x=0, will also have a point where f'(x) < 0. (This shows that even if f'(0) > 0, f is not increasing on any interval containing x=0"

2. Relevant equations

Well, this is part b of a problem. Part a was to find f'(0)=1, which I did by the mean-value theorem.

3. The attempt at a solution

If I could just get a hint on how to solve this, part b, I will attempt a solution.

2. Oct 15, 2009

### Dick

Write out f'(x) and think about what values it takes for x=1/(n*pi) where n is an integer.

3. Oct 15, 2009

### HallsofIvy

Staff Emeritus
How did you "use the mean value theorem" to find f'(0)= 0?

4. Oct 15, 2009

### Wingeer

Well, now you are making me unsure. But this is what I did:
$$f'(0)= \frac {f(1)-f(-1)}{1-(-1)} = \frac {(1+2 sin(1))-(-1 +2 sin(-1))}{2} = \frac {2+2 sin(1) - sin(-1)}{2} = \frac {2}{2} = 1$$

5. Oct 15, 2009

### Dick

Aside from being numerically wrong, that's not the definition of the derivative f'(0). Don't you have to take a limit to find f'(0)?

6. Oct 15, 2009

### Wingeer

Well. By the definition of the derivative, we have got $$f'(x) = {\lim}\limits_{h \to 0} \frac {x + 2(x+h)^2 \cdot sin( \frac{1}{x+h}) -(x + 2x^2 sin (\frac {1}{x}))}{h}$$

Edit: flaw in the expression. Thanks to Dick for pointing it out.

Last edited: Oct 15, 2009
7. Oct 15, 2009

### Dick

That getting there. But you want f' when x=0. That makes (f(x+h)-f(x))/h just (f(h)-f(0))/h. Since f(0)=0, that's just f(h)/h. BTW don't forget the x term in the definition of f(x).

8. Oct 15, 2009

### Wingeer

Did I forget something?
Yes, I see it now. The reason I didn't do this at first, was that I filled in, then substituted x with 0, and therefore got some undefined trigonometric identities.

Edit: Which I now see is defined by the definition of the function. My head is not top notch today.

Last edited: Oct 15, 2009
9. Oct 15, 2009

### HallsofIvy

Staff Emeritus
My point was that knowing that (f(1)- f(-1))/(1-(-1))= 1 only tells you that there is SOME x between -1 and 1 such that f'(p)= 1, not that that point is x=0!

The reason I asked instead of simply saying that the mean value theorem wouldn't work is that if you could find a sequence of points $\{x_n\}$ such that $(f(x_n)- f(-x_n))/(x_n- (-x_n))= 1$ then you could "squeez" that point to x= 0, but I couldn't see how to do that.

In any case, as Dick told you, the basic formula $f'(0)= lim (f(h)- f(0))/h$ works.

10. Oct 15, 2009

### Wingeer

Yes, I get it now. The second task is still a bit diffuse to me though.

11. Oct 15, 2009

### Dick

You are probably overlooking post 2.