f(x)={ 3-x  x<1
       { ax²+bx   x≥1

Find the unique values of a and b such that f(x) is continuous

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

so what i did is I used the definition of continuity:

f(x) is continuous at c IF

f(c)=lim_x→c-f(c)=lim_x→c+f(c)

a(1)²+b(1) = lim_x→1-f(c)=lim_x→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?

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.

Quote from: squidgetx on October 01, 2010, 08:25:29 pm

[ 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 Lim_x->c⁺f(x) = Lim_x->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.

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?

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.

Quote from: Deep Thought on October 02, 2010, 08:32:42 pm

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.