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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- anonymous

- katieb

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- anonymous

what is the function?

- anonymous

ok

- anonymous

Determine the intervals where the function is concave up and concave down.

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## More answers

- 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

for the first one the lowest point is at
y=-4

- anonymous

at f(0) i got that, isn't there a rule having to do with the second derivative i can
use?

- anonymous

Are you thinking of Leibniz's notation?

- 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

f' is equal to 0 when x is 2

- anonymous

- anonymous

oh i see

- anonymous

i got negative concavity for both sides of zero in [-2,3]

- anonymous

so does that mean it looks something like this? |dw:1367705711666:dw|

- anonymous

for f'... yes

- anonymous

what about f(x)?

- anonymous

the graph you drew is f(x)... sorry my bad!

- anonymous

f(x)' will look like this

- anonymous

I also need to label the inflection point and asymptotes. To find the inflection point i need to find f''(0) right?

- anonymous

|dw:1367706115115:dw|

- anonymous

what the..
how did you get that?

- anonymous

when it is down x is approx. -0.638 and y is approx. -3.222

- anonymous

Graphing calculator

- anonymous

and when it is up it is the same numbers... just positive

- anonymous

- anonymous

ahhh I see

- anonymous

so what does that tell us?

- anonymous

oh the slope! duh

- anonymous

Just type in the functions.
https://www.desmos.com/calculator
this is a good online graphing calculator

- anonymous

Thanks

- anonymous

np, tag me if you need me!

- anonymous

Thumps up! Im working on my practice final all afternoon so i might need you! @abarnett

- anonymous

ok I will be on as long as i can! what class are you taking?

- anonymous

Calc 1

- anonymous

cool

- 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

Yea I needed to completely work the problem beginning to end

- 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

@agent0smith ok, and that is really the only thing i need with the second derivative right?

- anonymous

|dw:1367707501528:dw|
very top y=10 and x=0

- agent0smith

Yep = it's just used to determine inflection points and concavity. Wherever f'' is neg, it's concave down, and vice versa.

- anonymous

that is f''

- anonymous

f'' just gives me inflection points at f'' = 0 and concavity.. cool! :)

- 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

@agent0smith you are right my bad! :P

- 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

@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

by the way im supposed to be using the interval [-2,3] for everything i forgot to tell you, haha

- anonymous

looks good to me, what do you think @agent0smith

- 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

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

ha! you got me, no infinities now!

- anonymous

so if there was no boundaries, the local and global maximum DNE right?

- 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

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

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

try this site it might help! never tried it
http://calculator.tutorvista.com/math/601/asymptote-calculator.html

- anonymous

- anonymous

I need to know how to do it from the ground up, no calculators

- 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

1!

- anonymous

so x^3/x^2 would have no horizontal, but have vertical?

- 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

and x^2/x^3 would have a horizontal of y=0?

- anonymous

no, and yes -1

- 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

i see so its like the graph of X

- anonymous

and if its negative im guessing thats what flips it over

- 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

excellent

- agent0smith

Slant asymptotes are a bit more confusing, sometimes you have to use polynomial long division to find them.

- anonymous

ok :) thank you so much, i wish i could give out 2 best answers

- anonymous

@Jgeurts I gave him one for you!

- anonymous

:) @abarnett

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