Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: squidgetx on October 01, 2010, 08:25:29 pm
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Today my calc teacher started off the class with a pop quiz, one of the questions of which has been a source of contention among my classmates and myself....
wow gotta lay off the ap english lol
anyway here is the problem:
f(x)={ 3-x x<1
{ ax2+bx x≥1
Find the unique values of a and b such that f(x) is continuous
so what i did is I used the definition of continuity:
f(x) is continuous at c IF
f(c)=limx→c-f(c)=limx→c+f(c)
a(1)2+b(1) = limx→1-f(c)=limx→1+f(c)
a(1)2+b(1)=3-1=a(1)2+b(1)
a+b=2
This is where I stopped. Some other people have told me that in order to find unique values, you have to make f(x) differentiable at 1 as well. I say that the question never says that that is necessary, so the correct answer is that there is no answer (or more accurately, any a and b such that a+b=2). Differentiability implies continuity, but continuity does not imply differentiability.
Who's right?
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Hmmm.....I haven't taken calculus yet (I'm in precalc/AP Stats) I would find the value of 3-x at 1, then set the second equation equal to that, and solve for one the variables. I guess that would get the same answer as you. Can't be much help, sorry.
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Today my calc teacher started off the class with a pop quiz, one of the questions of which has been a source of contention among my classmates and myself....
wow gotta lay off the ap english lol
anyway here is the problem:
f(x)={ 3-x x<1
{ ax2+bx x≥1
Find the unique values of a and b such that f(x) is continuous
so what i did is I used the definition of continuity:
f(x) is continuous at c IF
f(c)=limx→c-f(c)=limx→c+f(c)
a(1)2+b(1) = limx→1-f(c)=limx→1+f(c)
a(1)2+b(1)=3-1=a(1)2+b(1)
a+b=2
This is where I stopped. Some other people have told me that in order to find unique values, you have to make f(x) differentiable at 1 as well. I say that the question never says that that is necessary, so the correct answer is that there is no answer (or more accurately, any a and b such that a+b=2). Differentiability implies continuity, but continuity does not imply differentiability.
Who's right?
Hmm, seems like any value of A and B such that A+B=2 would work. Not sure what your teacher'd mean by "unique", though.
1024!
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well, not really.
The tangent lines for the two functions have to be the same. So, what you do is find the derivative of the functions. So:
f(x)={ -1 x<1
{ 2ax+b x≥1
which means that at 1, since the tangent lines need to be the same, and therefore the slope, and therefore the derivative, -1 = 2ax+b. So at x=1, -1=2a+b. This can be rearranged to get b=-1-2ax
So, at x=1, the two functions must equal each other to connect, as well as have their derivatives be the same. so 3-x=ax2+bx. See, now we're having fun. So, they must equal at x=1, so 3-1=a+b. or 2=a+b as you have pointed out.
Now we have two equations that equal b at x=1. So, 2=a-1-2a, or 2=-a-1, or -2=a+1, or a=-3. WHOA! So, with either of our equations, b=-1-2(-3), or b=-1+6=5. With the other one, 2=a+b, we have 2=-3+b, or b=5.
So, -3X2+5x is your second function.
f(x)={ 3-x x<1
{ -3x2+5x x≥1
The point at x=1 is (1,2) for both equations, and the tangent line is -1 for both equations. That wasn't too bad, was it?
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Well glad graphmastur got this. Without looking at my Calculus notes from last year all I could have told you is that you need to know what "a" and "b" equal. Without those being known you can't really solve it, or at least not at the calculus level :P
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The value of the function and the slope have to be equal at 1+ and 1-
3-x = ax2+bx
-1 = 2ax + b
Both occur at x=1 so:
2 = a + b
-1 = 2a + b
solving:
a = -3
b = 5
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Well my main issue was that I wasn't sure whether "continuous" means that the function is differentiable at that point. I say that it doesn't have to be differentiable, it just has to satisfy f(c)=limx->cf(c)
In general, we say that the function f is continuous at some point c of its domain if, and only if, the following holds:
The limit of f(x) as x approaches c through domain of f does exist and is equal to f(c); If the point c in the domain of f is not a limit point of the domain, then this condition is vacuously true, since x cannot approach c through values not equal c. Thus, for example, every function whose domain is the set of all integers is continuous.
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[ Differentiability implies continuity, but continuity does not imply differentiability.
Actually, continuity does imply differentiability. Let's define a function f(x) as continuous on some interval [a, b] if Limx->c+f(x) = Limx->c-f(x) =f(c) ∀ {c|a≤c≤b} This suggests that a change in c results in a finite change in f(c), which implies that the function has a finite rate of change for all points in the range [a, b]. Thus the derivative exists for the continuous function. It's not a mathematically rigorous proof, but you get the idea.
Note: I might actually know what I'm talking about or be completely wrong. I have no idea which one is right. Feel free to correct me.
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[ Differentiability implies continuity, but continuity does not imply differentiability.
Actually, continuity does imply differentiability. Let's define a function f(x) as continuous on some interval [a, b] if Limx->c+f(x) = Limx->c-f(x) =f(c) ∀ {c|a≤c≤b} This suggests that a change in c results in a finite change in f(c), which implies that the function has a finite rate of change for all points in the range [a, b]. Thus the derivative exists for the continuous function. It's not a mathematically rigorous proof, but you get the idea.
Note: I might actually know what I'm talking about or be completely wrong. I have no idea which one is right. Feel free to correct me.
Continuity only means that there are no asymptotes or holes. Differentiability, on the other hand, requires a continuous derivative (meaning no "corners"). So no, continuity would not imply differentiability.
well, not really.
The tangent lines for the two functions have to be the same. So, what you do is find the derivative of the functions. So:
f(x)={ -1 x<1
{ 2ax+b x≥1
which means that at 1, since the tangent lines need to be the same, and therefore the slope, and therefore the derivative, -1 = 2ax+b. So at x=1, -1=2a+b. This can be rearranged to get b=-1-2ax
So, at x=1, the two functions must equal each other to connect, as well as have their derivatives be the same. so 3-x=ax2+bx. See, now we're having fun. So, they must equal at x=1, so 3-1=a+b. or 2=a+b as you have pointed out.
Now we have two equations that equal b at x=1. So, 2=a-1-2a, or 2=-a-1, or -2=a+1, or a=-3. WHOA! So, with either of our equations, b=-1-2(-3), or b=-1+6=5. With the other one, 2=a+b, we have 2=-3+b, or b=5.
So, -3X2+5x is your second function.
f(x)={ 3-x x<1
{ -3x2+5x x≥1
The point at x=1 is (1,2) for both equations, and the tangent line is -1 for both equations. That wasn't too bad, was it?
The tangents don't have to be the same. It doesn't have to be differentiable, only continuous.
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Continuity only means that there are no asymptotes or holes. Differentiability, on the other hand, requires a continuous derivative (meaning no "corners"). So no, continuity would not imply differentiability.
http://mathworld.wolfram.com/ContinuousFunction.html (http://mathworld.wolfram.com/ContinuousFunction.html)
You'll notice that the more formal definition of a continuous function is equivalent to the definition I gave.
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That has nothing to do with the derivative, though. I think you're getting derivatives and limits mixed up :)
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Continuity does *not* imply differentiability. For example, abs(x) is continuous but not differentiable at x=0.
@OP: The quiz probably had a mistake - it probably wanted the unique pair (a,b) such that f is differentiable at x=1.
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Continuity does *not* imply differentiability. For example, abs(x) is continuous but not differentiable at x=0.
@OP: The quiz probably had a mistake - it probably wanted the unique pair (a,b) such that f is differentiable at x=1.
Exactly, and yeah, a mistake in the quiz would make sense.
EDIT: As clarification, Qwerty.55, the site says that the limit as X approaches any value A in a single-interval domain must exist on both sides for continuity, not the derivative. I think you're getting the two mixed up :)
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That has nothing to do with the derivative, though. I think you're getting derivatives and limits mixed up :)
That is always a probability.
And the abs(x) example caused me to see the error.
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That has nothing to do with the derivative, though. I think you're getting derivatives and limits mixed up :)
That is always a probability.
And the abs(x) example caused me to see the error.
Yeah, don't worry, I once got integrals and limits mixed up for no apparent reason. They have almost nothing in common x.x
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I would never have solved it if I knew this fight were to happen. Okay, so x implying y aside, the tangent lines must be the same for it to be continuos. This is because of the fact that as you approach from the right, eg the Ax2+Bx equation, we say the slope of the line is heading toward N. From the left, eg the 3-x equation, we say the slope of the line is heading toward M. So, for continuity, I'm sure that you would all agree that the two lines had to be connected at the point (1,2), since it was at x=1 and 3-x=2.
Okay, so I lied about the whole x not implying y thing. If it is differentiable, then it implies it is continuous. And yeah, the quiz was worded wrong I do believe. It should be differentiable not continuos. Either way, It doesn't matter. You are correct if it does not need to be differentiable. But, to make sure it is absolutely continuous, and getting exact answers, you should see if it is differentiable.
Wow, I just check recent-posts. This is gonna be ninja'd.
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I would never have solved it if I knew this fight were to happen. Okay, so x implying y aside, the tangent lines must be the same for it to be continuos. This is because of the fact that as you approach from the right, eg the Ax2+Bx equation, we say the slope of the line is heading toward N. From the left, eg the 3-x equation, we say the slope of the line is heading toward M. So, for continuity, I'm sure that you would all agree that the two lines had to be connected at the point (1,2), since it was at x=1 and 3-x=2.
Yes, they have to be connected, but they do not have to be connected with the same slope. In other words, tangent lines do not have to be the same at one point for it to be continuous. That's differentiability.
But anyway, you're right, squidgetx should check with his teacher to see if the question's written wrong.
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Yeah, I'll check it out. I'll let you know (hopefully on Monday)
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hm bit of a necropost
but anyway if anyone was interested, the question was worded wrong lol.
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That is fine. I'm glad you did though.