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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
 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
 one year ago

This Question is Closed

ovenmitt12Best ResponseYou've already chosen the best response.1
The vertex is (1,0)? and it has two other zeros? Or is it just a plain old parabola with the vertex at (1,1)?
 one year ago

EphiloBest 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
 one year ago

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

ovenmitt12Best ResponseYou've already chosen the best response.1
I 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....
 one year ago

ovenmitt12Best ResponseYou've already chosen the best response.1
My turn... What are the zeros of x^34x+1?
 one year ago

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

ovenmitt12Best ResponseYou've already chosen the best response.1
I 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.
 one year ago

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

ovenmitt12Best ResponseYou've already chosen the best response.1
dw:1363820293103:dw
 one year ago

ovenmitt12Best ResponseYou've already chosen the best response.1
lol sorry this is my first time, I didn't see the draw function.
 one year ago

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

ovenmitt12Best ResponseYou've already chosen the best response.1
Look at that... straight lines lol. I should have used those
 one year ago

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

ovenmitt12Best ResponseYou've already chosen the best response.1
dw:1363820563396:dw
 one year ago

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

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

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

ovenmitt12Best ResponseYou've already chosen the best response.1
Only if you answer mine lol
 one year ago

genius12Best ResponseYou've already chosen the best response.0
The 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
 one year ago

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

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

genius12Best ResponseYou've already chosen the best response.0
@Ephilo No medal no fan? lol =[
 one year ago

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

genius12Best ResponseYou've already chosen the best response.0
NOOO MEDALLLLLLLLL I WILLLL CRYYYYYYYYYYY
 one year ago

genius12Best ResponseYou've already chosen the best response.0
I JUST REMEMBERED U CANT GIVE ME A MEDAL LMFAO ITS OK
 one year ago

EphiloBest ResponseYou've already chosen the best response.0
lmaoooo 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
 one year ago

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

ovenmitt12Best ResponseYou've already chosen the best response.1
@Ephilo Zeros: x^34x+1 ?
 one year ago

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

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

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

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

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

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

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

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

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

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

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

ovenmitt12Best ResponseYou've already chosen the best response.1
yes that was my question... lol I know, delayed right
 one year ago

ovenmitt12Best ResponseYou've already chosen the best response.1
f ' (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.
 one year ago

ovenmitt12Best ResponseYou've already chosen the best response.1
Which ever gives you the highest and lowest values are your extremas.
 one year ago

ovenmitt12Best ResponseYou've already chosen the best response.1
so 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
 one year ago

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

SandeepReddyBest ResponseYou've already chosen the best response.0
Example 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).
 one year ago

EphiloBest 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
 one year ago

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

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

SandeepReddyBest ResponseYou've already chosen the best response.0
A 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
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

EphiloBest ResponseYou've already chosen the best response.0
Given 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.
 one year ago

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

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

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

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

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

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

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