Math Is Fun Forum

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Peacewise and Beyond

Let f be the function defined as:

f(x) = { x+2 , x < 2
{ ax^2+bx, x >(or equal to) 2

a) what is the relationship between a and b?
I got 2a+b=2
b) find the unique values of a and b that will make f both continuous and differentiable.

i substitued (2-2a) in to b and got down to y=ax^2+2x-2ax, but now what?

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: Peacewise and Beyond

Hi,

Could you show me how you got your answer to the first part?

I don't see that any particular relationship is implied between a and b. All we are told is that if x >= 2 then f(x) = ax^2+bx but this alone doesn't impose a relationship between a and b.

For the second part however, if we need f(x) to be everywhere continuous and differentiable, then we had better make sure that at x = 2 both x + 2 and ax^2 + bx are equal, so we have:

Call this equation (1).

If this condition is met, then the function will be everywhere continuous and may possibly be everywhere differentiable (but not necessarily*), but there are no unique values of a and b implied here, for instance:

a = 2 & b = -2 will satisfy equation (1) as will a = 3 & b = -4 as will an infinite number of other paris of values.

* If a function is not continuous at a point, then there is no tangent line and the function is not differentiable at that point. However, even if a function is continuous at a point, it may not be differentiable there. For example, the function  y = |x| is continuous at x=0, but it is not differentiable there, due to the fact that the limit in the definition of differentiation does not exist (the limit from the right is 1 while the limit from the left is −1). Graphically, we see this as a "kink" in the graph at x=0. Thus, differentiability implies continuity, but not vice versa. One famous example of a function that is continuous everywhere but differentiable nowhere is the Weierstrass function - http://en.wikipedia.org/wiki/Weierstrass_function

Last edited by gnitsuk (2006-11-23 05:04:03)

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Re: Peacewise and Beyond

the 2a+b=2 is apparently the relationship of a and b according to our teacher. He wants you to make the parabola so that the very tip of it is connected to the x+2

gnitsuk

Member

Registered: 2006-02-09

Posts: 121

Re: Peacewise and Beyond

Ok.

Here's the full answer. I think the question is wrongly worded. There should be no part a) and b). Part b) alone is enough. I would have written it thus:

Let f be the function defined as:

f(x) = { x+2 , x < 2
          { ax^2+bx, x >= 2

Find the unique values of a and b that will make f both continuous and differentiable.

Now, we have to make sure that the y values of the two functions are equal when x = 2 - this will ensure the graph is everywhere continuous. We did this is the previous post and got:

Call this equation (a)

Lastly, we need to make sure that the gradients of the two functions are equal at x = 2 - this will ensure differentiability.

So we need to set the differential of the LHS equal to the differential of the RHS at x = 2 giving:

Call this equation (b)

So we solve (a) and (b) simultaineously to give

and

So that the function required is:

f(x) = { x+2 , x < 2
          { -0.5x^2+3x, x >= 2

To see the function go here:

http://mathdemos.gcsu.edu/mathdemos/pie … ility.html

Scroll down to the "Example 2" applet and then delete the second function listed there and for the first function type in (x<=2)?x+2:-0.5*x^2+3*x and then press the "New Functions" button.

Last edited by gnitsuk (2006-11-23 22:21:24)