on limits and continuity theory?

Question

show that the function F(x) = [(x-a)^2 * (a-b)^2] +x takes on the value (a+b)/2 for some value of x.

Answer 1

Well....... It's a simple quadratic. Set it equal to (a+b)/2 and complete the math to get it in standard quadratic form, then show the discriminant is ≥ 0 and you have your proof.

Doug

Answer 2

F(a) = 0 + a = a
F(b) = 0 + b = b

Let's say that a a < (a+b)/2 < b (It's the same kind of if b < a. And, if a = b, then (a+b)/2 = a and it's already proven).

So, (a+b)/2 belongs to the interval [a,b] = [f(a), f(b)] (Remembr that f(a) = a and f(b) = b) and F is a continous function, so, it exist a c that belongs to the interval (a,b) so that f(c) = (a+b)/2

Perhaps it would be clearer if you do the graphic of a similar situation. It doesn't matter what happens in the "middle" of [a,b], the function can grow or be constant or decrease n times. But the important part is that, if f is a continous function then you can state that *at least* once it covers every values between f(a) and f(b). And, since f(a) = a and f(b) = b in this case, and (a+b)/2 is a value inbetween, then there will be at least a c so that f(c) = (a+b)/2

...... ...... |
.......... b |..................o.. .......... ............o
............. |................o. o....... ...............o
(a+b)/2. |..............o.....o.. ...............o
..............|......................i.o..........o.... i
...........a |............o........i....o.....o...... i
............. | ......................i.........o........ l
............. |____ _____ i____ _____i______
...........................a .......c'.. ........... ....c.. b

Things are different if f is not a continous function

Example:

2 |......... ____
1 | ____
..|___________
.......a.........b

As you can see, there are many values k between f(a) = 1 and f(b) = 2 so that there is not a c that belongs to (a,b) and so that f(c) = k. For instance, there is not a c in (a,b) so that f(c) = 1.5, there is not a c in (a,b) so that f(c) = 1.8, and so on, and so on... As they say: ad infinitum.


Ilusion