calc help (finding values such that the function is cont)

[paulxzt]

Hey could someone help me with these, im stuck on some:

1) Determine the values of b and c such that the function is continuous for all reals.

f(x) = x + 1 , 1 < x < 3

x^2 + bx + c, x< 1 and x > 3

my attempt: f(1) = b+c and f(3) = 9 + 3b + c. Am i doing this completely wrong? Please help.

2) Determine the intervals on which the function is continuous.

f(x) = 5 - x, x<2
2x - 3 , x > 2

Thanks

 

[Unco]

Re: calc help (finding values such that the function is cont

1) Determine the values of b and c such that the function is continuous for all reals.

f(x) = x + 1 , 1 < x < 3  
       x^2 + bx + c, x<=1 and x>=3

my attempt: f(1) = b+c and f(3) = 9 + 3b + c. Am i doing this completely wrong? Please help.

2) Determine the intervals on which the function is continuous.

f(x) = 5 - x, x<=2
       2x - 3 , x > 2

1) Continuity of f(x) at x=a requires that

\(\displaystyle \L \mbox{\lim_{x\to a^+} f(x) = f(a) = \lim_{x\to a^-} f(x)}\)

2) It should be clear that f(x) is continous for x<2 and x>2, so we only need to think about x=2. Does the limit of f(x) as x approach 2 from the left equal that from the right? Graphically, do the two lines meet at x=2?

 

[merlin2007]

Re:Calc Help

a) Since each part of the function is continuous by itself, we just need to make sure that the parabola has the same values as the line on the borderlines.

At x = 1, the line = 1+1=2
We require that the parabola, which is 1+b+c at x=1, also =2
Thus 1+b+c = 2, so b+c = 1

At x=3, the line = 3+1 = 4
We require that the parabola, which is 9+3b +c at x=3, also =4
Thus, 9+3b +c = 4, and 3b+c = -5

We now have two equations in two variables. Solving yields b= -3, c=4

b) As in the previous problem, each part is continuous in and of itself. We just have to check whether it's continuous at the borderline.
5-2 = 3
2(2) - 3 = 1

Since it is not continuous at the border, but is continuous elsewhere, we say that the intervals on which the function is continuous is:

(-Infinity, 2) and (2, Infinity)

These two open intervals indicate that x=2 is the only non-continuous point.


(Note, I rarely do online tutoring, but I felt I should give back, since I've gotten help here before. If someone wants to critique my solutions, please do so.)

 

[soroban]

Re: calc help (finding values such that the function is cont

Hello, paulxzt!

1) Determine the values of \(\displaystyle b\) and \(\displaystyle c\) such that the function is continuous for all reals.

. . \(\displaystyle f(x) \:=\;\left\{\begin{array}{cc}x\,+\,1& \quad & 1\,<\,x\,<\,3 \\ x^2 \,+\,bx\,+\,c & & x\,\leq\,1,\;x\;\geq \,3\end{array}\)

My attempt: \(\displaystyle \,f(1) \:= \:b\,+\,c\,\) and \(\displaystyle \,f(3) \:= \:9\,+\,3b\,+\.c\).
Am i doing this completely wrong?

You are on the right track . . .

The middle portion of the function looks like this:

            |
            |           o(3,4)
            |         * :
            |       *   :
            |     *     :
            |   o(1,2)  :
            |   :       :
            |   :       :
      ------+---+-------+--
            |     1         3

We want the parabola to "join up" with the line segment.

So: \(\displaystyle \,f(1)\:=\:1^2\,+\,b\cdot1\,+\,c\:=\;2\;\;\Rightarrow\;\;b\,+\,c\:=\:1\;\) [1]

And: \(\displaystyle \,f(3) \:=\:3^2\,+\,b\cdot3\,+\,c\:=\:4\;\;\Rightarrow\;\;3b\,+\,c\:=\:-5
\;\) [2]

Subtract [1] from [2]: \(\displaystyle \:2b\:=\:-6\;\;\Rightarrow\;\;\fbox{b\,=\,-3}\)

Substitute into [1]: \(\displaystyle \:-3\,+\,c\:=\:1\;\;\Rightarrow\;\;\fbox{c\,=\,4}\)

2) Determine the intervals on which the function is continuous.
. . . \(\displaystyle f(x) \:= \:\left\{\begin{array}{cc}5\,-\,x & \quad & x\,\leq\, 2 \\ 2x\,-\,3 & & x\,>\,2\end{array}\)

      \ |            /
        \           /
        | \        /
        |   \     /
        |     *  /
        |       /
        |      /
        |     o
        |     :
  ------+-----+----------
        |     2
        |

\(\displaystyle f(x)\) is continuous for all \(\displaystyle x\,\neq\,2:\;\;\fbox{(-\infty,\,2)\,\cup\,(2,\,\infty)}\)