Expert: Josh - 12/18/2008

Question
QUESTION: Hi,
You did such a great job with my last question that I am going to ask you one more. Thanks so much.

Let F(x) be the function defined below:
   
    [(x^2)-4xk+3h , x>or=4
F(x)=[k , -2<x<4
    [(3/2)((x^3)k)-3xh , x<or=-2

Find the values of k and h so that f(x) is continuous everywhere. Justify your answer.

ANSWER: The conditions of continuity require
f'(x=-2-eps) = f'(x=-2+eps)
f'(x=4-eps)  = f'(x=4+eps)
in the limit as "eps" goes to 0.

In plain English, the curves must be joined seamlessly.

Since f(x) is constant in the interval (-2,4), f'(x=-2+eps) = f'(x=4-eps) must in fact be zero.

The first derivatives are
(9/2)*k*x^2-3h  for x<=-2
0          for -2<x<4
2x-4k          for x>=4

Evaluating f'(x) at x=-2 and x=4, respectively, then equating each to zero yields the solution. If I'm not mistaken,

k=x/2 (evaluate at x=4)
h=(3k/2)x^2 (evaluate at x=-2)

What do you think?

---------- FOLLOW-UP ----------

QUESTION: Sorry, I have a few follow up questions. Firstly, what does "eps" mean or stand for? Second, k and h have to be constants. I do not quite understand what is going on here. If you could, please explain this answer to me a little more. Sorry for not understanding, and thanks so much again.

Tippy

Answer
That's all right. "eps" is the Greek alphabet epsilon, which you can look up on a search engine. It represents a tiny quantity and is frequently used in calculus when we consider the value of a function in the limit, for instance, as eps goes to zero. It is the same sort of thing when we looked at asymptotes last time, approaching a limiting value "xo" from the right hand side (we consider f(x) at xo+eps) and from the left hand side (we consider f(x) at xo-eps). It refers to the value of the function just to the left (respectively, right) of the point xo.

I misread the question in my earlier reply. We don't need to guarantee smoothness, so we don't need to differentiate at all. To obtain a continuous function, we just need the boundary points to be connected.

For a function to be continuous at the points xo=-2 and xo=4, the value of the function in a neighborhood around the point must have the same value as the function AT the point. Since the function has only one variable, we require f(xo)=f(xo-eps)=f(xo+eps) in the limiting case as "eps" shrinks to zero when we converge on the point.

Let's look at each of the quantities,
Just left of xo=-2, at x=-2-eps
f(-2-eps)=(3k/2)*(-8-12*eps-6*eps^2-eps^3)+6h+3h*eps
Just right of xo=-2, at x=-2+eps
f(-2+eps)=k
In the limit as eps->0, f(-2-eps)=f(-2+eps), with all "eps" terms tending to zero, after simplifying, we obtain
-12k+6h=k.
The first condition to be satisfied is 6h-13k=0 ...[1]

Just right of xo=4, at x=-4+eps
f(4+eps)=(16+8*eps+eps^2)-4k(4+eps)+3h
       =16+3h-16k+4*(2-k)*eps+eps^2
Equating f(4+eps) to f(4-eps) in the limit as eps->0,
16+3h-16k=k
The second condition is 3h-17k=-16 ...[2]

Solving these two equations [1] and [2] simultaneously,
[steps omitted]
k=32/21, h=416/126

I hope this makes sense now. Thanks for the follow-up question.