General Discussion > Math and Science
math stuffs
squidgetx:
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:
--- Quote ---f(x)={ 3-x x<1
{ ax2+bx x≥1
Find the unique values of a and b such that f(x) is continuous
--- End quote ---
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?
MRide:
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.
Deep Toaster:
--- Quote from: squidgetx on October 01, 2010, 08:25:29 pm ---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:
--- Quote ---f(x)={ 3-x x<1
{ ax2+bx x≥1
Find the unique values of a and b such that f(x) is continuous
--- End quote ---
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?
--- End quote ---
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!
jnesselr:
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?
meishe91:
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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