anonymous
  • anonymous
Below you are given a function f(x) and its first and second derivatives. Use this information to solve the following.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
what is the function?
anonymous
  • anonymous
ok
anonymous
  • anonymous
Determine the intervals where the function is concave up and concave down.

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anonymous
  • anonymous
\[f(x) = (x^2 - 4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(1-3x^2)/(x^2 +1)^3\]
anonymous
  • anonymous
for the first one the lowest point is at y=-4
anonymous
  • anonymous
at f(0) i got that, isn't there a rule having to do with the second derivative i can use?
anonymous
  • anonymous
Are you thinking of Leibniz's notation?
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
f' is equal to 0 when x is 2
anonymous
  • anonymous
@Jgeurts
anonymous
  • anonymous
oh i see
anonymous
  • anonymous
i got negative concavity for both sides of zero in [-2,3]
anonymous
  • anonymous
so does that mean it looks something like this? |dw:1367705711666:dw|
anonymous
  • anonymous
for f'... yes
anonymous
  • anonymous
what about f(x)?
anonymous
  • anonymous
the graph you drew is f(x)... sorry my bad!
anonymous
  • anonymous
f(x)' will look like this
anonymous
  • anonymous
I also need to label the inflection point and asymptotes. To find the inflection point i need to find f''(0) right?
anonymous
  • anonymous
|dw:1367706115115:dw|
anonymous
  • anonymous
what the.. how did you get that?
anonymous
  • anonymous
when it is down x is approx. -0.638 and y is approx. -3.222
anonymous
  • anonymous
Graphing calculator
anonymous
  • anonymous
and when it is up it is the same numbers... just positive
anonymous
  • anonymous
@Jgeurts
anonymous
  • anonymous
ahhh I see
anonymous
  • anonymous
so what does that tell us?
anonymous
  • anonymous
oh the slope! duh
anonymous
  • anonymous
Just type in the functions. https://www.desmos.com/calculator this is a good online graphing calculator
anonymous
  • anonymous
Thanks
anonymous
  • anonymous
np, tag me if you need me!
anonymous
  • anonymous
Thumps up! Im working on my practice final all afternoon so i might need you! @abarnett
anonymous
  • anonymous
ok I will be on as long as i can! what class are you taking?
anonymous
  • anonymous
Calc 1
anonymous
  • anonymous
cool
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
Yea I needed to completely work the problem beginning to end
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
@agent0smith ok, and that is really the only thing i need with the second derivative right?
anonymous
  • anonymous
|dw:1367707501528:dw| very top y=10 and x=0
agent0smith
  • agent0smith
Yep = it's just used to determine inflection points and concavity. Wherever f'' is neg, it's concave down, and vice versa.
anonymous
  • anonymous
that is f''
anonymous
  • anonymous
f'' just gives me inflection points at f'' = 0 and concavity.. cool! :)
agent0smith
  • agent0smith
@Abarnett probably easier to just link to one than hand draw a graph :P f'': https://www.google.com/search?q=10(1-3x%5E2)%2F(x%5E2+%2B1)%5E3&aq=f&oq=10(1-3x%5E2)%2F(x%5E2+%2B1)%5E3&aqs=chrome.0.57j60l3j62l2.174j0&sourceid=chrome&ie=UTF-8
anonymous
  • anonymous
@agent0smith you are right my bad! :P
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
@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
anonymous
  • anonymous
by the way im supposed to be using the interval [-2,3] for everything i forgot to tell you, haha
anonymous
  • anonymous
looks good to me, what do you think @agent0smith
anonymous
  • anonymous
\[f(x) = (x^2 - 4)/(x^2 +1)\]\[f'(x) = 10x/(x^2 +1)^2\]\[f''(x) = 10(1-3x^2)/(x^2 +1)^3\]
agent0smith
  • agent0smith
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=UTF-8 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] ;)
anonymous
  • anonymous
ha! you got me, no infinities now!
anonymous
  • anonymous
so if there was no boundaries, the local and global maximum DNE right?
agent0smith
  • agent0smith
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.
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
try this site it might help! never tried it http://calculator.tutorvista.com/math/601/asymptote-calculator.html
anonymous
  • anonymous
@Jgeurts
anonymous
  • anonymous
I need to know how to do it from the ground up, no calculators
agent0smith
  • agent0smith
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 ...?
anonymous
  • anonymous
1!
anonymous
  • anonymous
so x^3/x^2 would have no horizontal, but have vertical?
agent0smith
  • agent0smith
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? \[\Large f(x) = \frac{ (x^2 - 4)}{(x +1)}\]
anonymous
  • anonymous
and x^2/x^3 would have a horizontal of y=0?
anonymous
  • anonymous
no, and yes -1
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
i see so its like the graph of X
anonymous
  • anonymous
and if its negative im guessing thats what flips it over
agent0smith
  • agent0smith
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.
anonymous
  • anonymous
excellent
agent0smith
  • agent0smith
Slant asymptotes are a bit more confusing, sometimes you have to use polynomial long division to find them.
anonymous
  • anonymous
ok :) thank you so much, i wish i could give out 2 best answers
anonymous
  • anonymous
@Jgeurts I gave him one for you!
anonymous
  • anonymous
:) @abarnett

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