Derivatives problem 17.mws

1. Rolle's Theorem and the Mean Value Theorem.

According to the statement of the Mean Value Theorem, if f is a function which is continuous on the closed interval [ a, b ] , which is differentiable on the interior of this interval, then there is some c in the open interval ( a, b ) such that  D(f)(c) = (f(b)-f(a))/(b-a) .  The proof of the theorem involves considering the secant line between the points ( a, f(a) ) and ( b, f(b) ), which is defined as a function by

> g:=x-> (f(b)-f(a))/(b-a)*(x-a)+f(a);

g := proc (x) options operator, arrow; (f(b)-f(a))/...

Then the new function h defined by h(x) = f(x)-g(x) satisfies the hypotheses of Rolle's theorem, so that there is some c in ( a, b ) such that D(h)(c) = 0 . This value of c satisfies the conclusion of the Mean Value Theorem for the function f . Let us illustrate this proof by examining an example. Suppose we consider the function f below on the interval [-1, 2].

> f:=x->x^3-2*x^2+1;

f := proc (x) options operator, arrow; x^3-2*x^2+1 ...

> a:=-1;b:=2;

a := -1

b := 2

> g(x);

x-1

> h:=f-g;

h := f-g

> h(x);

x^3-2*x^2+2-x

> h(-1);h(2);

0

0

Since h(-1) = 0 and h(2) = 0 , Rolle's Theorem applies, and there is some c in ( -1, 2 ) such that D(h)(c) = 0 . Let us plot the function to see this.

> plot(h,-1..2);

[Maple Plot]

In fact, we see that there are 2 such values. We can find them explicitly as follows.

> solve(D(h)(x)=0);

2/3+1/3*sqrt(7), 2/3-1/3*sqrt(7)

Now, we could also have simply solved the original Mean Value problem as follows:

> solve(D(f)(x)=(f(b)-f(a))/(b-a));

2/3+1/3*sqrt(7), 2/3-1/3*sqrt(7)

Let us plot the function, its secant line, and the two tangent lines to see that they are parallel.

> T:=x->D(f)(c)*(x-c)+f(c);

T := proc (x) options operator, arrow; D(f)(c)*(x-c...

> c:=2/3-1/3*sqrt(7);T(x);

c := 2/3-1/3*sqrt(7)

(3*(2/3-1/3*sqrt(7))^2-8/3+4/3*sqrt(7))*(x-2/3+1/3*...

> c:=2/3+1/3*sqrt(7);T(x);

c := 2/3+1/3*sqrt(7)

(3*(2/3+1/3*sqrt(7))^2-8/3-4/3*sqrt(7))*(x-2/3-1/3*...

> plot([f(x),g(x),(3*(2/3-1/3*sqrt(7))^2-8/3+4/3*sqrt(7))*(x-2/3+1/3*sqrt(7))+(2/3-1/3*sqrt(7))^3-2*(2/3-1/3*sqrt(7))^2+1,(3*(2/3+1/3*sqrt(7))^2-8/3-4/3*sqrt(7))*(x-2/3-1/3*sqrt(7))+(2/3+1/3*sqrt(7))^3-2*(2/3+1/3*sqrt(7))^2+1],x=-1..2,color=[red,blue,green,orange]);

[Maple Plot]

The picture makes it very clear that the tangent lines at these points are parallel to the secant line.

Submission:

For the function given by f(x) = x*(sin(x)+x) ,

(a) Plot the function on the interval [ 0, 2*Pi ] . Estimate the values of c which satisfies the conclusion of the mean value theorem. Then compute them exactly.

(b) Define the function g whose graph is the secant line through ( 0, f(0) ) and ( 2*Pi, f(2*Pi) ) and plot f-g . Estimate the values of c which satisfy the conclusion of Rolle's theorem for f-g . Then compute them exactly.

(c) What is the relationship between your answer for part a) and part b)?

(d) Plot f, g, and any tangent lines through points (c,f(c)) which satisfy the conclusion of the mean value theorem. What about your picture illustrates the mean value theorem?

Submission worksheet: