Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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So... what exactly is this assignment? it looks tricky.

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Omg which one????

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Number 0! I don't get it!!!! It says to compute limits and determine continuity and differentiability of the function in a graph.

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Here's the picture of the graph.

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ok to begin with limits helps to study the trend of a function near an input (a value). And determining continuity and differentiability helps to see how the function acts and how its values change near a point.
Let's start on how to find the limit of a point, which are presented as $lim f(x)$ as $x->1$
which is read as "as the limit of f(x) approaches 1, what is the y value?"

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Some problems with the same value have a minus or possitive sign to the right of the number, what does that mean?

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Sometimes in graphs a value of f(x), may have two different outputs and be written as follows,
$lim f(x)$ as $x->1^-$
$lim f(x)$ as $x->1^+$
$lim f(x)$ as $x->1$
When the number has a negative sign it reads as "what is the limit of f(x) as it approaches from the right", and when it has a positive sign, it reads as "what is the limit of f(x) as it approaches from the left"
When doesn't have any signs (right or left), just follow the point to tell if it has a limit, but a good way to tell if $lim f(x)$ as $x->1$ has a limit, if your left and right limits of the same value are the same, than the limit of $lim f(x)$ as $x->1$ exists, but if you get two different numbers, than the value is undefined, here I'll show you.

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See how we got two different value for the limit of f(x) at 1?

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oh ok! So...
$lim f(x)$ as $x->1^-$ = 4
$lim f(x)$ as $x->1^+$ = -2
$lim f(x)$ as $x->1$ = UND

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Yes!!!!! Keep going.
Here are some rules to help you out
- a fuction $lim f(x)$ as $x->a$ if only
$lim f(x)$ as $x->a^-$ equals the same of $lim f(x)$ as $x->a^+$

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Ok this is what I answered for the review question.
$lim f(x)$ as $x->-4$ = 2
$lim f(x)$ as $x->1$ = UND
$lim f(x)$ as $x->6$ = 5

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Yes, that is correct! How do you know what value to choose if there are to points at one value of f(x), such f(6)?

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Here's what I did


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The next question is "Identify all values of a such that "$lim f(x)$ as $x->a$ is undefined.
According to my graph it looks like it's only f(1), because the ponts are disconnected with two different lines. The limit comes from both sides, like you said so it is UND.

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Correct!

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Next problem reads as "Identify all values of a such that "$f(x)$ is undefined at $x = a$"

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That concept is very simple, if you remember from Pre-Calculus I, when there is an open circle a any value of x for y, then it means that it is undefined.

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Ok so the answer to values of a such that "$f(x)$ is undefined at $x = a$"
is $x = 4$
The others have undefined circles too, but I see that they have two values (one from the right and the other from the left, and this point only has one.

Next questions is "Identify all value of a such that $f(x)$ is discontinous at $x = a$"
I don't understand it thaty well, can you help?

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Yes, here are some good rules, if the point doesn't meet this criteria, than it is not continous

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1) a function f(x) is continous at a pount x = a. f(x) has a limit x -> a. f(x) is defined at x = a, and if both values have the same lim (left and right). If the limit of f(x) equals the limit of f(a). Here is an easy visual.


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Ok, so the line of graph literally gets cut off, than it is discontinous (hence the name), and the circles, if you think about it, add a break in between the line, so it makes it discontinous.
So the asnwer to "Identify all value of a such that $f(x)$ is discontinous at $x = a$" are x = -4, 1, 6 ?

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YES!

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The last question goes as follows "Identify all values of a such that f(x) is not differentiable at x = a"
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Here is a good way to determine differentiablity.

Basically you don't want, sharp points, disconnected lines, open circles, sharp vertical lines, or lines that go up and down in a short interval (ex q).

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So all values of a such that f(x) is not differentiable at $x = a$ are x = -4, 1, 6

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Yes, I'm so proud of you!!!

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That was so easy!!! I don't know how I ever messed it up! Thank you!