page 186 section 4.15 #2

Use Rolle's Theorem to prove that regardless of the value of b , there is at most one point x in the interval -1..1 for which .

Suppose there were two such points . Then by Rolles Theorem, there would be a value of c between and such that . But a simple inspection shows that the derivative is zero only at .

> f:=x->x^3-3*x+b;

> solve(f(-1)=0,b);solve(f(1)=0,b);

With some thought, we see that there is one solution in the interval [-1,1] if and only if b is between -2 and 2.

> with(plots):

> animate(x^3-3*x+b,x=-1..1,b=-3..3,thickness=3);

page 186 section 4.15 #3

> f:=x->piecewise(x<=1,(3-x^2)/2,1/x);

> plot(f(x),x=0..2,thickness=3);

In order to show that f satifies the hypotheses of the mean value theorem in the interval [0,2] it suffices to show that f is continuous and differentiable at .

> limit(f(x),x=1,left);limit(f(x),x=1,right);f(1);

> limit(D(f)(x),x=1,left);limit(D(f)(x),x=1,right);

> solve(f(2)-f(1)=D(f)(c)*(2-1),c);

> plot([f(x),f(1/2)+D(f)(1/2)*(x-1/2),f(sqrt(2))+D(f)(sqrt(2))*(x-sqrt(2))],x=0..2,color=[red,blue,blue],thickness=[3,2,2]);

page 186 section 4.15 #8(a)

Use the mean value theorem to show that for all x,y .

For any x,y by the mean value theorem there is a c between x and y such that . Now for any c we have so the inequality follows.

This property says that the sine function is a "contraction". The distance between is less than or equal to the distance betwee x and y

> A:=plot3d(abs(sin(y)-sin(x)),x=-Pi..Pi,y=-Pi..Pi,axes=boxed,style=patchcontour):

> B:=plot3d(abs(y-x),x=-Pi..Pi,y=-Pi..Pi,axes=boxed,style=wireframe,thickness=2):

> display(A,B);

The contractive property suggests that domains of the interates of the sine function contract. In the graph below, note how the range is shrinking.

> plot([seq((sin@@n)(x),n=1..50)],x=-Pi..Pi,color=[red,green,blue]);

>