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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?
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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)
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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
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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
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)
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