I finished the following problems I'm not sure if I have done them correctly so I need someone to tell me if they are correct.
13 is wrong because you should have the dot on the lower function to show that it includes x=2 on that range. Likewise, in the upper function there should be no dot to show that it is not continuous on x=2. Furthermore, you should include the dot at x=6 and show that it is at x=6 by doing what you did with x=2. For 14 f(3) for 3-x = 0 and (x-3)^2 is 0 as well. This implies that it is continuous at x=3. It is not continuous on x=0 because sqrt(-x) at 0 is 0 while 3-x at zero is 3. This means it jumps at x=3 to three. A good way to know if a piecewise function is continuous is by checking the values of all endpoints for the two common functions.
Thanks for taking your time to help me out! For 13, I am really confused as to what I'm supposed to do. I read your comment like a hundred times but I still don't get it. Would it be possible for you to sketch the graph because I'm having a difficult time understanding it. As for 14, I understand it now so thanks. I updated the picture so could you let me know if it's correct this time both #13 and #14?
hey, check out the videos on www.mathvideoguy.com about continuity. if that doesn't help. pm me and ill help you
Here's the deal with continuity:
A function $\displaystyle f$ that's defined on an interval $\displaystyle (x_0,x_1)$ is continuous at $\displaystyle x=a$, $\displaystyle a \in (x_0,x_1)$,
holds precisely when the following 3 things are true:
1) $\displaystyle a$ is in the domain of $\displaystyle f$ (i.e. $\displaystyle f(a)$ is defined)
(The way I wrote this, that's already given in my setup $\displaystyle f$ defined on an interval $\displaystyle (x_0,x_1)$, and $\displaystyle a \in (x_0, x_1)$.)
2) $\displaystyle \lim_{x \rightarrow a} f(x)$ exists.
3) $\displaystyle \lim_{x \rightarrow a} f(x) = f(a)$.
That's often summarized as just #3: $\displaystyle \lim_{x \rightarrow a} f(x) = f(a)$, because in that statement, #1 and #2 are implicitly required.
For #2, the $\displaystyle \lim_{x \rightarrow a} f(x)$ exists, holds precisely when the following 3 things are true:
2.1) $\displaystyle \lim_{x \rightarrow a^-} f(x)$ exists.
2.2) $\displaystyle \lim_{x \rightarrow a^+} f(x)$ exists.
2.3) $\displaystyle \lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x)$
If 2.1, 2.2, and 2.3 hold, then $\displaystyle \lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x)$.
You could put these together to give a new list of requirements for $\displaystyle f$ to be continous at $\displaystyle x = a$:
1') $\displaystyle a$ is in the domain of $\displaystyle f$ (i.e. $\displaystyle f(a)$ is defined)
2') $\displaystyle \lim_{x \rightarrow a^-} f(x)$ exists.
3') $\displaystyle \lim_{x \rightarrow a^-} f(x) = f(a)$.
4') $\displaystyle \lim_{x \rightarrow a^+} f(x)$ exists.
5') $\displaystyle \lim_{x \rightarrow a^+} f(x) = f(a)$.
Or, since the limit exists and $\displaystyle a$ is in the domain are implicit in these statements, as just these two conditions:
1'') $\displaystyle \lim_{x \rightarrow a^-} f(x) = f(a)$.
2'') $\displaystyle \lim_{x \rightarrow a^+} f(x) = f(a)$.
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For the case of an endpoint, when $\displaystyle f$ is defined on an interval $\displaystyle (x_0, a]$, f is continuous at $\displaystyle x=a$ exactly when both:
1) $\displaystyle a$ is in the domain of $\displaystyle f$ (i.e. $\displaystyle f(a)$ is defined)
(The way I wrote this, that's already given in my setup $\displaystyle f$ defined on an interval $\displaystyle (x_0,a]$.)
2) $\displaystyle \lim_{x \rightarrow a^-} f(x)$ exists.
3) $\displaystyle \lim_{x \rightarrow a^-} f(x) = f(a)$.
Again. we could simplfy this to $\displaystyle \lim_{x \rightarrow a^-} f(x) = f(a)$.
The other, lower endpoint works the same.
The issue with continuity at the endpoints is that the function simply isn't know to exist outside that interval, so only one of the "one-sided limits" is even meaningful.
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Ex: #12: it's continuous when a=2. Reason:
#1' holds: $\displaystyle x = 2$ is in the domain of $\displaystyle f$. ($\displaystyle f(2) = (2)^3 = 8$.)
#2' holds: $\displaystyle \lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} (x^3)$ exists.
#3' holds: $\displaystyle \lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} (x^3) = 2^3 = 8 = f(2)$.
4') holds: $\displaystyle \lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^+} 2x^2$ exists.
5') holds: $\displaystyle \lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^+} 2x^2 = 2(2^2) = 8 = f(2)$.
Therefore $\displaystyle f$ is continous at $\displaystyle x = 2$.
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Ex: #13: Your graphed function is continuous at x = 2 on the interval $\displaystyle [0,2]$. Reason:
Only need show that $\displaystyle \lim_{x \rightarrow 2^-} f(x) = f(2)$.
But, from the graph, $\displaystyle f(x) = 2$ when $\displaystyle x \in [0,2]$, thus
$\displaystyle \lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} (2) = 2 = f(2)$.
BE AWARE, it's correct to say "f is continuous at x = 2 on [0,2]". But f is NOT continuous at x = 2 on [0, 6].
When you talk about "f on [0,2]", you pretend that that's all there is to the function, and so don't look at x values outside that domain.
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Ex: #14: f is not continuous at x = 2. Reason:
2'' holds: $\displaystyle 0$ is in the domain $\displaystyle (f(0) = 3 - 0 = 3)$ and $\displaystyle \lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} (3-x) = 3 = f(0)$.
1'') Does not hold: $\displaystyle \lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \sqrt{-x} = 0 \ne f(0)$.
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Ex: #14: f is not continuous at x = 3. Reason:
$\displaystyle \lim_{x \rightarrow 3^-} f(x) = \lim_{x \rightarrow 3^-} (3-x) = 0$, and $\displaystyle \lim_{x \rightarrow 3^+} f(x) = \lim_{x \rightarrow 3^+} (x-3)^2 = 0$.
Therefore $\displaystyle \lim_{x \rightarrow 3^-} f(x) = \lim_{x \rightarrow 3^+} f(x) = 0$, and so $\displaystyle \lim_{x \rightarrow 3} f(x)$ exists and equals $\displaystyle 0$.
But $\displaystyle f$ is not continous at $\displaystyle 0$, because 0 is not in the domain of $\displaystyle f$.
It's impossible that $\displaystyle \lim_{x \rightarrow 3} f(x) = f(3)$, because $\displaystyle f(3)$ is not defined.
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