A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
can some one graph this please??
ill award and fan !
it is an upsidedown parabola with a vertex at x=1 (third charateristic) and zeros at x=0 and x=2 (first charateristic) and the second says that the slope of the tangent line is positive when x<1 so the graph rises from left to right until it gets to x=1 and the fourth says that the slope is negative meaning it falls from left to right after x=1 making the graph an upsidedown parabola
anonymous
 3 years ago
can some one graph this please?? ill award and fan ! it is an upsidedown parabola with a vertex at x=1 (third charateristic) and zeros at x=0 and x=2 (first charateristic) and the second says that the slope of the tangent line is positive when x<1 so the graph rises from left to right until it gets to x=1 and the fourth says that the slope is negative meaning it falls from left to right after x=1 making the graph an upsidedown parabola

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The vertex is (1,0)? and it has two other zeros? Or is it just a plain old parabola with the vertex at (1,1)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 the original question is this Sketch the graph of a function f(x) having the following characteristics...? f (0) = f (2) = 0 f ' (x) > 0 if x < 1 f ' (1) = 0 f ' (x) < 0 if x > 1 f " (x) < 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought that would be what the graph would look like

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think you have it.... It would be a parabola with: vertex (1,1) P(0,0) and P(2,0) and with f''(x) < 0 it would be concave down....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0My turn... What are the zeros of x^34x+1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i understand it i just don't write get how im suppose to graph that :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think you just have to graph the original function that would have been a parabola... I don't think you would have to graph any of the derivative or the second deriv.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you draw that out please,because what the graph is suppose to look like is were im lost

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363820293103:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol sorry this is my first time, I didn't see the draw function.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363820649485:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Look at that... straight lines lol. I should have used those

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats funny lol but thanks it really helped!! :P :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363820563396:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 thanks so much !

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363821010459:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 can you help me with another question please?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Only if you answer mine lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The equation becomes: f(x) = x(x  2) = x^2 + 2x If the tangents to graph are positive for x < 1, then that just means the parabola is increasing from (infinity, 1). For it to be increasing over that interval, it needs to be upside down, and so I placed a negative sign in front of the leading coefficient, which reflects it in the xaxis. Secondly, if the graph has a vertex at x = 1, then that implies this: f'(1) = 0 > Mean the value of the first derivative at x = 1 is 0. Let's see if that's the case. f'(x) = 2x + 2 > f'(1) = 2(1) + 2 = 0 > It works! But we need to know what the yvalue of the vertex is at x = 1. The derivative tells us that there definitely is a vertex at x = 1, now to find the yvalue of this vertex, we just plug in x = 1 in to our function and get the value: f(1) = (1)(1  2) = (1)(1) = 1 > So the vertex is at (1, 1) So this equations gives us that there is definitely a vertex at x = 1, moreover, at (1, 1), it's increasing for x < 1, and it's zeroes are at x = 0 and x = 2. But because we need to graph it, why not also find the yintercept so it's easier? And we know that the yintercept occurs when x = 0. So we plug in 0 for x in the function and get y = 0. So the yintercept occurs at y = 0. This means that both our x and y intercepts occur at (0, 0). So that's everything. Now we just graph with the information we have. dw:1363820753206:dw @ephilo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@genius12 WOW! thanks thats perfect!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 what would that be?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Ephilo No medal no fan? lol =[

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@genius12 i did fan you ,lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0NOOO MEDALLLLLLLLL I WILLLL CRYYYYYYYYYYY

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I JUST REMEMBERED U CANT GIVE ME A MEDAL LMFAO ITS OK

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lmaoooo i cant give everyone a medal but if i could i swear i could that was a lot of help,btw i actually did read it @genius12

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@genius12 ill give you medal if you help me on my other question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Ephilo Zeros: x^34x+1 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol okay i have 5 more that i dont understand out of 15 questions i guess thats good :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 whats that for? is that your question ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0find the absolute extrema and the xvalue where they occur on the given interval f(x)x^312x [0,4]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that f(x) = x^312x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363822851080:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0absolute extrema is 16 and the x value is x =4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SandeepReddy wow that was fast im still trying to understand the question :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ephilo do u want me to explain?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please.. @SandeepReddy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the one i wrote as answer was absolute maximum. absolute minimum also exists. at x = 0, the and the value is 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0two absolute extrema's one minimum and one maximum

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Coming to the clarification part

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes that was my question... lol I know, delayed right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f ' (x) = 3x^2  12 critical numbers: 3(x^24) = 0 x = 2 x = 2 Plug those back into the original. Also plug the 0 and 4 from your interval into the original.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Which ever gives you the highest and lowest values are your extremas.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so f (2) = 16 and both f (2) and f (4) give you 16 x value of 2 gives extrema 16 ; x values 2 and 4 give you 16

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 and @SandeepReddy thanks guys ! :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Example 1 Determine the absolute extrema for the function g(t) = 2t^3+3t^212t+4 and interval.[4,2] Solution All we really need to do here is follow the procedure given below 1) Find all critical points of f(x) that are in the interval [a,b]. This makes sense if you think about it. Since we are only interested in what the function is doing in this interval we don’t care about critical points that fall outside the interval. Evaluate the function at the critical points found in step 1 and the end points. Identify the absolute extrema. Now, we need to get the derivative so that we can find the critical points of the function. g'(t) = 6t^2+6t12 and make it equals to zero and find t's It looks like we’ll have two critical points, and . Note that we actually want something more than just the critical points. We only want the critical points of the function that lie in the interval in question. Both of these do fall in the interval as so we will use both of them. That may seem like a silly thing to mention at this point, but it is often forgotten, usually when it becomes important, and so we will mention it at every opportunity to make sure it’s not forgotten. Now we evaluate the function at the critical points and the end points of the interval. Absolute extrema are the largest and smallest the function will ever be and these four points represent the only places in the interval where the absolute extrema can occur. So, from this list we see that the absolute maximum of g(t) is 24 and it occurs at (a critical point) and the absolute minimum of g(t) is 28 which occurs at (an endpoint).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SandeepReddy thanks that helped a lot, i have two more its okay if you cant help tho youve helped a lot

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay if you say so @SandeepReddy find any critical numbers of f(x)=4x/x^2+1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363825177046:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A number a in the domain of a given function f is called a critical number of f if f '(a) = 0 or f ' is undefined at x = a

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363825539582:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363825649852:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363825701530:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363825784023:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SandeepReddy yes, i was just reading over it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol @SandeepReddy you tired yet?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Until I get the medal LOL

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I wont get tired hahahahah

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohhh you want a medal lol okay !!@SandeepReddy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol yeah sadly, im so sleepy tho.sigh but this is due at 9 so id better get it done

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Given the derivative, f ‘(x) = 6x2 + 12x a.State the critical values where f ‘(x) =0 b. Find the second derivative, f “(x) c. Use the second derivative to test the critical values found in part a. Analyze the results to determine whether these critical values produce a relative maximum or minimum.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a) 6x2 = 12x 6x212x = 6x(x2) =0 => x=0 and x = 2 are critical points

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0plug the values of crtical points we got in a) and check whether they produce minima or maxima

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SandeepReddy okay i understand that one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes 2 freaking more !!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0apply the mean value theorem to f on [0,1] to find all values of c such that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1363826906752:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hey ask that last question. i have to go

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SandeepReddy some ones already answering it , thanks for your help though i really appreciate it :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Your Welcome and im leaving, have a good day, good luck with studies!!
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.