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

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Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1what is the function?

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1Determine the intervals where the function is concave up and concave down.

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x) = (x^2  4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(13x^2)/(x^2 +1)^3\]

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1for the first one the lowest point is at y=4

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

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1Are you thinking of Leibniz's notation?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Use 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.

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1f' is equal to 0 when x is 2

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1i got negative concavity for both sides of zero in [2,3]

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1so does that mean it looks something like this? dw:1367705711666:dw

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1the graph you drew is f(x)... sorry my bad!

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1f(x)' will look like this

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

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1dw:1367706115115:dw

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1what the.. how did you get that?

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1when it is down x is approx. 0.638 and y is approx. 3.222

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1and when it is up it is the same numbers... just positive

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1so what does that tell us?

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

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1np, tag me if you need me!

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

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1ok I will be on as long as i can! what class are you taking?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Did 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.

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1Yea I needed to completely work the problem beginning to end

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Once 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.

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

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1dw:1367707501528:dw very top y=10 and x=0

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Yep = it's just used to determine inflection points and concavity. Wherever f'' is neg, it's concave down, and vice versa.

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1f'' just gives me inflection points at f'' = 0 and concavity.. cool! :)

agent0smith
 one year ago
Best 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

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1@agent0smith you are right my bad! :P

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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.

Jgeurts
 one year ago
Best 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

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

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1looks good to me, what do you think @agent0smith

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x) = (x^2  4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(13x^2)/(x^2 +1)^3\]

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1lol 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] ;)

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1ha! you got me, no infinities now!

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1so if there was no boundaries, the local and global maximum DNE right?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1If 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.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Correct, 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.

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1Ok, 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?

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

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1I need to know how to do it from the ground up, no calculators

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1To 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 ...?

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1so x^3/x^2 would have no horizontal, but have vertical?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Correct! :) 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? \[\Large f(x) = \frac{ (x^2  4)}{(x +1)}\]

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1and x^2/x^3 would have a horizontal of y=0?

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1For 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.

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1i see so its like the graph of X

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1and if its negative im guessing thats what flips it over

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Yep. 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.

agent0smith
 one year ago
Best ResponseYou've already chosen the best response.1Slant asymptotes are a bit more confusing, sometimes you have to use polynomial long division to find them.

Jgeurts
 one year ago
Best ResponseYou've already chosen the best response.1ok :) thank you so much, i wish i could give out 2 best answers

Abarnett
 one year ago
Best ResponseYou've already chosen the best response.1@Jgeurts I gave him one for you!
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