A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Are these correct?
anonymous
 one year ago
Are these correct?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How is that one correct? I got the same answers as the first 3 problems but not number 4.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe your goal is to plug in x, the x is stated for you. (2)^2 + 5(2) + 6  (2)+2 4+10+6  4 20/4 = 5. Doing this in my head, apologies for any error.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I made a mistake in that. Thanks! Can you explain number 5 too @Mikeyy1992?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I shall try, one sec.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x1  x^2  1 X is stated to be 1. 11  1^2  1 0  1*1  1 0  0 The answer shall be 0, if I am not mistaken.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay I got that too but my teacher put 1/2.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you know how to solve 6 and 7? Sorry if I'm asking for too much.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can do them in my head, but I do not know how to fully explain 6 and 7, but from my knowledge, the answers are both correct. For 5, I have dealt with the same exact puzzle, so I know the answer is 0, anyways.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you write the steps on how you solve it? I'll see if I can understand when I look at the steps.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.06, already proves it will be DNE, with the second statement of x>2. With number 7, I see cos, so already think negative wise, from past learnings, which I know by heart. It's more so, making e to e then x/3 into a factor which produces e^3. I am not good at explaining some terms in mathematics, apologies.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why does the second statement x>2 prove that it's DNE?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Zepdrix, might point you in a better direction than me; if I am making any mistakes.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oooo it's Princess Jasmin!! :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Are these answers that the teacher posted? Or are we checking YOUR answers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No it's my teacher's answers and haha ya that's me! :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2For number 6, remember that for a limit to exist `at a particular point`, The limits from each direction must match \(\large\rm x\to2^{+}\) needs to give the same as \(\large\rm x\to2^{}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Gosh, thank you for stating that way better than me, zepdrix! Knew you could! (:

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \lim_{x\to2^{}}2x^24x\]So if you look at this big ole piecewise function... when we approach 2 from the `left side`, we're using the top piece, understand why? :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The top piece meaning 2x^2 4x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We are trying to see if the function is indeterminate or if the point doesn't exist? Or if the function is continuous or if it has a hole?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2For piecewise functions, we need to make sure that the `limit from the left` agrees with the `limit from the right`. If they don't agree, then the limit does not exist at that point. So we have some big piecewise function, let's call it f(x). We want to know whether or not \(\large\rm \lim_{x\to2}f(x)\) exists, and if so, find the value. To determine this, we need to check and see if the left and right limits agree,\[\large\rm \lim_{x\to2^}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So again I'll ask,\[\large\rm \lim_{x\to2^{}}f(x)=\lim_{x\to2^{}}2x^24x\]Do you understand why I'm replacing f(x) with this top line in the piecewise function for x approaching from the left?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so wouldn't you just insert in a 2 on both sides of the equation? It would still give the same answer right?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes, if that's easier, maybe just think of it that way :) The pieces need to give the same output at x=2.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 2x^24x \stackrel{?}{=}4\sin\left(\frac{\pi x}{4}\right)\]So are these equal when you plug in x=2? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So no it doesn't match.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh so do you that process for all inequalities that pop up with the lim function?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2These were our steps: \[\large\rm \lim_{x\to2^}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]\[\large\rm \lim_{x\to2^}2x^24x\stackrel{?}{=}\lim_{x\to2^+}4\sin\left(\frac{\pi x}{4}\right)\]\[\large\rm 2(2)^24(2)\stackrel{?}{=}4\sin\left(\frac{4\pi}{4}\right)\]\[\large\rm 0\ne4\]I suppose yes. It might get a tiny bit more complicated if your teacher throws piecewise functions at you with 3 or 4 pieces. But it should always be the same process :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Should we go over number 5? :o looks like you were both a little confused on that one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ya can you go over that and number 7? Sorry!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \lim_{x \to1}\frac{x1}{x^21}=\frac{0}{0}\ne0\]0/0 is an indeterminate form, it's not equal to 0. So we have to work our way around it some how.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Notice that the denominator is the difference of squares,\[\large\rm x^21=x^21^2\]We have a rule for factoring the difference of squares, do you remember it? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we have to multiply both top and bottom by x+1?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Umm sure, that's another option! That's pretty much the same thing I was doing, but in reverse.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay. That's how my teacher taught me. So after reducing or crossing out you would get 1/(x+1)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And then you would have to plug in 1 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, that's what I forgot, been a couple years! :P Thank you again, Zepdrix.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhhh that's why the answer is 1/2!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So this is what you do with limits. Step 1: You try to plug the number directly in. ~If you run into trouble, you have to try and `cancel stuff out`. Step 2: Cancel stuff out. Step 3: Try to plug the number directly in again.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay I understood that. Thanks! Can you help me with number 7?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2See the steps I listed? Number 7 is going to be a piece of cake for us. We'll apply step 1, and we won't run into any problems. It will simplify down to an answer.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Plug 3 directly into the function. We get something like this,\[\large\rm e^{3}\cos\left(\frac{\pi(3)}{3}\right)\]yes?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[e^3\cos(3.14) = e^3(1) = e^3\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2You and your ways _ lol You gotta remember your trig stuff silly billy! \(\large\rm \cos(\pi)=1\) < yes. I don't like the 3.14 though lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol sorry! But thank you soo much!! You are really good at helping others! :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2yay team \c:/ we did it!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @Mikeyy1992! You were very helpful too! :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.