1. ## Differentiability Question?

1. Suppose f(x) = x2 sin(1/x)+x/2 if x is not 0 and defne f(0) = 0:
(a) Show that f'(0) > 0:
(b) Show that no matter how small the positive number h may be, there are infnitely
many points on both sides of x = 0 and within a distance of h of x = 0 at which
f'(x) = 3/2 and also infnitely many at which f(x) = -1/2 : This shows that there is
no interval, in particular no neighborhood Vh(0), on which f is increasing, in spite
of the fact that the slope of the curve y = f(x) is positive at 0.
2. Let f be differentiable when a < x < b: Suppose x1 and x2 are distinct points of the
interval, and let P1 and P2 be the corresponding points of the curve y = f(x):
(a) Show that the condition for P1 to lie above that line tangent to the curve at P2 is
f(x1)-f(x2) - (x1-x2)f'(x2) > 0:
(b) If the condition in (a) holds for every pair of points on the curve, show that f'(x)
increases as x increases. In fact, if x1 < x2, show that
f'(x1) <(f(x2) -f(x1))/(x2 -x1)< f'(x2):

I just get stuck starting.
Like I don't know if I use the MVTtheorem or what?

2. f'(0)=lim[f(x)-f(0)/x-c]=lim[x^2sin(1/x)+x/2-0/x]=lim[x^2sin(1/x)+x/2/x]=lim[xsin(1/x)+1/2]=1/2
Thus f'(0)>0

Now b has me completely confused on what it is even asking...

3. Originally Posted by mathematic
Now b has me completely confused on what it is even asking...

We have

$\displaystyle f'(x)=2x\sin \dfrac{1}{x}-\cos \dfrac{1}{x}-\dfrac{1}{2}\quad (x\neq 0)$

Denote

$\displaystyle x_k=\dfrac{1}{2k\pi}\;,\;y_k=\dfrac{1}{(2k+1)\pi}$

then

$\displaystyle f'\left(x_k\right)=-\dfrac{1}{2}\;,\;f'\left(y_k\right)=\dfrac{3}{2}\q uad (\forall k\in \mathbb{Z}-\{0\})$

and

$\displaystyle \displaystyle\lim_{k \to{\pm}\infty}{x_k}=0\;,\;\displaystyle\lim_{k \to{\pm}\infty}{y_k}=0$

Now, you can conclude.

Fernando Revilla

4. Ok, is it valid to just find sequences like that so it is true for the proof?

Now for 2.
a)f is differentiable, so it must be continuous.
x1 and x2 are in the interval, so we can apply the MVT.
f(x1)-f(x2)/x1-x2
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