Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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

Calculus1
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

what is the function?
ok
Determine the intervals where the function is concave up and concave down.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question