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can some one graph this please?? ill award and fan ! it is an upside-down 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 upside-down parabola

Mathematics
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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)?
@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
i thought that would be what the graph would look like

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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....
My turn... What are the zeros of x^3-4x+1?
i understand it i just don't write get how im suppose to graph that :(
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.
can you draw that out please,because what the graph is suppose to look like is were im lost
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lol sorry this is my first time, I didn't see the draw function.
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Look at that... straight lines lol. I should have used those
lol!
lolololol!
thats funny lol but thanks it really helped!! :P :D
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@ovenmitt12 thanks so much !
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@ovenmitt12 can you help me with another question please?
Only if you answer mine lol
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 x-axis. 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 y-value of the vertex is at x = 1. The derivative tells us that there definitely is a vertex at x = 1, now to find the y-value 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 y-intercept so it's easier? And we know that the y-intercept occurs when x = 0. So we plug in 0 for x in the function and get y = 0. So the y-intercept 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
@genius12 WOW! thanks thats perfect!
@ovenmitt12 what would that be?
@Ephilo No medal no fan? lol =[
@genius12 i did fan you ,lol
NOOO MEDALLLLLLLLL I WILLLL CRYYYYYYYYYYY
I JUST REMEMBERED U CANT GIVE ME A MEDAL LMFAO ITS OK
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
@genius12 ill give you medal if you help me on my other question
@Ephilo Zeros: x^3-4x+1 ?
Ephilo ask me
i mean ask us! :)
lol okay i have 5 more that i dont understand out of 15 questions i guess thats good :/
@ovenmitt12 whats that for? is that your question ?
find the absolute extrema and the x-value where they occur on the given interval f(x)x^3-12x [0,4]
is that f(x) = x^3-12x
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absolute extrema is 16 and the x value is x =4
@SandeepReddy wow that was fast im still trying to understand the question :(
Ephilo do u want me to explain?
please.. @SandeepReddy
yeah wait
the one i wrote as answer was absolute maximum. absolute minimum also exists. at x = 0, the and the value is 0
two absolute extrema's one minimum and one maximum
Coming to the clarification part
yes that was my question... lol I know, delayed right
f ' (x) = 3x^2 - 12 critical numbers: 3(x^2-4) = 0 x = 2 x = -2 Plug those back into the original. Also plug the 0 and 4 from your interval into the original.
Which ever gives you the highest and lowest values are your extremas.
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
@ovenmitt12 and @SandeepReddy thanks guys ! :D
Example 1 Determine the absolute extrema for the function g(t) = 2t^3+3t^2-12t+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+6t-12 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).
@SandeepReddy thanks that helped a lot, i have two more its okay if you cant help tho youve helped a lot
Its OKay
ask me!
ask me no problem!
okay if you say so @SandeepReddy find any critical numbers of f(x)=4x/x^2+1
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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
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x = +1 or -1
@Ephilo Is it clear?
@SandeepReddy yes, i was just reading over it.
Okay!
lol @SandeepReddy you tired yet?
Nope !
Until I get the medal LOL
I wont get tired hahahahah
ohhh you want a medal lol okay !!@SandeepReddy
hey i m kidding~
u have any question?
lol yeah sadly, im so sleepy tho.sigh but this is due at 9 so id better get it done
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.
a) 6x2 = 12x 6x2-12x = 6x(x-2) =0 => x=0 and x = 2 are critical points
b) its -12x+12
plug the values of crtical points we got in a) and check whether they produce minima or maxima
@SandeepReddy okay i understand that one
hmm good!
yes 2 freaking more !!
shoot
lol what ?
ask!
okay okay lol
apply the mean value theorem to f on [0,1] to find all values of c such that
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https://www.khanacademy.org/math/calculus/derivative_applications/mean_value_theorem/v/mean-value-theorem
Hey ask that last question. i have to go
@SandeepReddy some ones already answering it , thanks for your help though i really appreciate it :)
Your Welcome and im leaving, have a good day, good luck with studies!!

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