Differentiability and continuity.   

p. 99, #6.

>	Q:=t->piecewise(t<=0,1,t>0,exp(-t));

>	plot(Q(t),t=-2..2);

Q(t) appears to be a continuous function of t since there is no break in the graph.  

However, it appears that the graph has a corner at t=0 which suggests that Q(t) is not differentiable there.  

This is the same as saying that I(t)=Q'(t)=dQ/dt is not defined at t=0.  

In order to see this more clearly we will compute some difference quotients first for negative h.

>	(Q(-.1)-Q(0))/(-.1-0);

>	(Q(-.01)-Q(0))/(-.01-0);

No surprise.  The graph is flat if t<0.  Now for positive h.

>	(Q(.1)-Q(0))/(.1-0);

>	(Q(.05)-Q(0))/(.05-0);

>	(Q(.01)-Q(0))/(.01-0);

>	(Q(.005)-Q(0))/(.005-0);

>	(Q(.001)-Q(0))/(.001-0);

What do you guess the limit to be?

>	diff(Q(t),t);

p. 99, #10.

>	f:=y->piecewise(y>=0,sqrt(y^2+1)-y,y<0,sqrt(y^2+1)+y);

>	plot(f(y),y=-2..2);

The function appears to be continuous at y=0 but not differentiable there.  We can verify this by looking at some di fference quotients.

>	(f(-.1)-f(0))/(-.1-0);

>	(f(-.03)-f(0))/(-.03-0);

>	(f(-.01)-f(0))/(-.01-0);

>	(f(-.003)-f(0))/(-.003-0);

>	(f(-.001)-f(0))/(-.001-0);

What do you guess the limit to be?

>	(f(.1)-f(0))/(.1-0);

>	(f(.05)-f(0))/(.05-0);

>	(f(.01)-f(0))/(.01-0);

>	(f(.005)-f(0))/(.005-0);

>	(f(.001)-f(0))/(.001-0);

What do you guess the limit to be?  

Final example.

>	g:=x->piecewise(x<=1,3*x-2,x>1,x^3);

>	plot(g(x),x=0..2);

This function is definitely continuous and it appears that it might be differentiable as well.  Let's check some difference quotients.

>	(g(.9)-g(1))/(.9-1);

>	(g(.95)-g(1))/(.95-1);

No surprise here.  The function is linear with slope 3 for x<=1.

>	(g(1.1)-g(1))/(1.1-1);

>	(g(1.05)-g(1))/(1.05-1);

>	(g(1.01)-g(1))/(1.01-1);

>	(g(1.005)-g(1))/(1.005-1);

>	(g(1.001)-g(1))/(1.001-1);

What is your guess for the limit?