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 one year 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
 one year 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

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ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1The vertex is (1,0)? and it has two other zeros? Or is it just a plain old parabola with the vertex at (1,1)?

Ephilo
 one year 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

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0i thought that would be what the graph would look like

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1I 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....

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1My turn... What are the zeros of x^34x+1?

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

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

Ephilo
 one year 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

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1dw:1363820293103:dw

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1lol sorry this is my first time, I didn't see the draw function.

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363820649485:dw

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1Look at that... straight lines lol. I should have used those

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

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1dw:1363820563396:dw

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 thanks so much !

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363821010459:dw

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 can you help me with another question please?

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1Only if you answer mine lol

genius12
 one year 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

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@genius12 WOW! thanks thats perfect!

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 what would that be?

genius12
 one year ago
Best ResponseYou've already chosen the best response.0@Ephilo No medal no fan? lol =[

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@genius12 i did fan you ,lol

genius12
 one year ago
Best ResponseYou've already chosen the best response.0NOOO MEDALLLLLLLLL I WILLLL CRYYYYYYYYYYY

genius12
 one year ago
Best ResponseYou've already chosen the best response.0I JUST REMEMBERED U CANT GIVE ME A MEDAL LMFAO ITS OK

Ephilo
 one year 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

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

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1@Ephilo Zeros: x^34x+1 ?

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0i mean ask us! :)

Ephilo
 one year 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 :/

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 whats that for? is that your question ?

Ephilo
 one year 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]

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0is that f(x) = x^312x

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0absolute extrema is 16 and the x value is x =4

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

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0Ephilo do u want me to explain?

SandeepReddy
 one year 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

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

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0Coming to the clarification part

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1yes that was my question... lol I know, delayed right

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1f ' (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.

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1Which ever gives you the highest and lowest values are your extremas.

ovenmitt12
 one year ago
Best ResponseYou've already chosen the best response.1so 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

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@ovenmitt12 and @SandeepReddy thanks guys ! :D

SandeepReddy
 one year 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).

Ephilo
 one year 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

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0ask me no problem!

Ephilo
 one year 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

SandeepReddy
 one year 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

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363825539582:dw

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363825649852:dw

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363825701530:dw

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0dw:1363825784023:dw

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0@Ephilo Is it clear?

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@SandeepReddy yes, i was just reading over it.

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0lol @SandeepReddy you tired yet?

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0Until I get the medal LOL

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0I wont get tired hahahahah

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0ohhh you want a medal lol okay !!@SandeepReddy

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0u have any question?

Ephilo
 one year 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

Ephilo
 one year 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.

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

SandeepReddy
 one year 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

Ephilo
 one year ago
Best ResponseYou've already chosen the best response.0@SandeepReddy okay i understand that one

Ephilo
 one year 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

SandeepReddy
 one year ago
Best ResponseYou've already chosen the best response.0Hey ask that last question. i have to go

Ephilo
 one year 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 :)

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