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Below you are given a function f(x) and its first and second derivatives. Use this information to solve the following.
 11 months ago
 11 months ago
Below you are given a function f(x) and its first and second derivatives. Use this information to solve the following.
 11 months ago
 11 months ago

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AbarnettBest ResponseYou've already chosen the best response.1
what is the function?
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
Determine the intervals where the function is concave up and concave down.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
\[f(x) = (x^2  4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(13x^2)/(x^2 +1)^3\]
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
for the first one the lowest point is at y=4
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
at f(0) i got that, isn't there a rule having to do with the second derivative i can use?
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
Are you thinking of Leibniz's notation?
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Use the second derivative to find concavity. It's concave down where f'' is negative, ie f'' < 0 Concave up where f'' is positive, f'' > 0 Find where f'' = 0 firstly.
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
f' is equal to 0 when x is 2
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
i got negative concavity for both sides of zero in [2,3]
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
so does that mean it looks something like this? dw:1367705711666:dw
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
the graph you drew is f(x)... sorry my bad!
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
f(x)' will look like this
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
I also need to label the inflection point and asymptotes. To find the inflection point i need to find f''(0) right?
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
dw:1367706115115:dw
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
what the.. how did you get that?
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
when it is down x is approx. 0.638 and y is approx. 3.222
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
and when it is up it is the same numbers... just positive
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
so what does that tell us?
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
Just type in the functions. https://www.desmos.com/calculator this is a good online graphing calculator
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
np, tag me if you need me!
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
Thumps up! Im working on my practice final all afternoon so i might need you! @abarnett
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
ok I will be on as long as i can! what class are you taking?
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Did you find f''? once you do, you can find where it's concave up and down. Just from looking at the graph, I'd guess concave down between about inf and 1, and +1 and +infinity Concave up between about 1 and 1.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
Yea I needed to completely work the problem beginning to end
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Once you get the x values where f'' = 0, pick points to the left and right of those x values, and check if f'' is positive (concave down) or negative (concave up) to help find the intervals.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
@agent0smith ok, and that is really the only thing i need with the second derivative right?
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
dw:1367707501528:dw very top y=10 and x=0
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Yep = it's just used to determine inflection points and concavity. Wherever f'' is neg, it's concave down, and vice versa.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
f'' just gives me inflection points at f'' = 0 and concavity.. cool! :)
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
@Abarnett probably easier to just link to one than hand draw a graph :P f'': https://www.google.com/search?q=10(13x%5E2)%2F(x%5E2+%2B1)%5E3&aq=f&oq=10(13x%5E2)%2F(x%5E2+%2B1)%5E3&aqs=chrome.0.57j60l3j62l2.174j0&sourceid=chrome&ie=UTF8
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
@agent0smith you are right my bad! :P
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Oh and @Jgeurts f'' can also be used to find if a critical point (where f' = 0) is a max or min  at a maximum, f'' is negative. At a minimum, f'' is positive. But that's part of the concavity anyway, since maximums are concave down, minimums are concave up. If f'' is zero at a point where f' = 0, then it's an inflection point.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
@Abarnett @agent0smith So I have Global max (3,1/2), Global Min (0,4), Decreasing on the interval [2,0], Increasing on the interval [0,3], local max is f(3), local min is f(0), concave down at [inf,sqrt1/3]U[sqrt1/3, inf], concave up at [sqrt1/3, sqrt1/3], inflection points occur at +sqrt1/3
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
by the way im supposed to be using the interval [2,3] for everything i forgot to tell you, haha
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
looks good to me, what do you think @agent0smith
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
\[f(x) = (x^2  4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(13x^2)/(x^2 +1)^3\]
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
lol that interval [2,3] makes a big difference... as the original function has no maximum, it just has an asymptote. https://www.google.com/search?q=(x%5E2++4)%2F(x%5E2+%2B1)&aq=f&oq=(x%5E2++4)%2F(x%5E2+%2B1)&aqs=chrome.0.57j60l3j62l2.269j0&sourceid=chrome&ie=UTF8 That's the original function... all your values look good. Except: concave down at [inf,sqrt1/3]U[sqrt1/3, inf], Something seems off here, with an interval [2,3] ;)
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
ha! you got me, no infinities now!
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
so if there was no boundaries, the local and global maximum DNE right?
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
If you can learn to kinda read the graph, you can check your values. The max/mins should be easy to find (turning points), same with where the graph is increasing or decreasing (the slope f' is positive or negative). Points of inflection are more difficult to see... you kinda have to gauge where it changes from concave up to down which can be hard to see, but still possible to estimate where it is.
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Correct, since f' will never equal 0 except at the minimum. \[\large f'(x) = 10x/(x^2 +1)^2\]  note that this can be only zero if x=0 (the denominator can never make the f' equal zero)  ie there's no other turning points/maximums, only a minimum.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
Ok, i feel so informed, haha. Hold up, last question! What about asymptotes? Horizontal/Vertical? Obviously theres one around y=1 but how do i prove it?
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
try this site it might help! never tried it http://calculator.tutorvista.com/math/601/asymptotecalculator.html
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
I need to know how to do it from the ground up, no calculators
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
To find asymptotes... you just have to look at the original function, there's no calculus involved. \[\large f(x) = (x^2  4)/(x^2 +1)\] What happens when x gets really huge (positive or negative infinity? The 4 and +1 become tiny compared to the huge x^2, so it becomes \[\Large \frac{ x^2 }{ x^2 }\] which is equal to ...?
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
so x^3/x^2 would have no horizontal, but have vertical?
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Correct! :) that's for horizontal asymptotes. So y= 1 is a horizontal asymptote. To find vertical asymptotes, find values of x that cause the denominator to be zero \[\Large f(x) = \frac{ (x^2  4)}{(x^2 +1)}\]are there any x values that make the denominator zero? x^2 is always positive, and you're adding 1... What about instead for this function? \[\Large f(x) = \frac{ (x^2  4)}{(x +1)}\]
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
and x^2/x^3 would have a horizontal of y=0?
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
For x^3/x^2, that simplifies to just x  it's actually a slant asymptote (neither horizontal or vertical)  there is no horizontal. BUT x can't equal zero, so it has a vertical at x=0 (or just a 'hole' at x=0, not really an asymptote in this case), since the denominator is zero.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
i see so its like the graph of X
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
and if its negative im guessing thats what flips it over
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Yep. With a hole at x=0. So for vertical  look for values of x that give a denominator zero For horizontal  look for how the function behaves when x is approaching +infinity or infinity.
 11 months ago

agent0smithBest ResponseYou've already chosen the best response.1
Slant asymptotes are a bit more confusing, sometimes you have to use polynomial long division to find them.
 11 months ago

JgeurtsBest ResponseYou've already chosen the best response.1
ok :) thank you so much, i wish i could give out 2 best answers
 11 months ago

AbarnettBest ResponseYou've already chosen the best response.1
@Jgeurts I gave him one for you!
 11 months ago
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