Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
HELP PLEASE?!
use the Second Derivative Test to find all
relative extrema.
 one year ago
 one year ago
HELP PLEASE?! use the Second Derivative Test to find all relative extrema.
 one year ago
 one year ago

This Question is Closed

satellite73Best ResponseYou've already chosen the best response.4
\[h(t)=t4\sqrt{t+1}\]?
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
i guess you need the first two derivatives what did you get for the first one?
 one year ago

EphiloBest ResponseYou've already chosen the best response.0
@satellite73 give me a second im gonna do it right now
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
yes it is i would write it as \[1\frac{2}{\sqrt{t+1}}\] but that is correct.
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
so before finding the second derivative, we need the critical points one critical point is where the derivative is undefined, at \(t=1\) which is also the left hand endpoint of the domain of \(h\) since the domain of \(h\) is \([1,\infty)\) to find the other set \[1\frac{2}{\sqrt{t+1}}=0\] and solve for \(t\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
you are asked to use the second derivative to find the relative max and min of your function that means you are asked to use the second derivative to test the critical points to see if they are max or min or neither in order to do that, you need to find the critical points first
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
critical points where the derivative is zero or where the derivative is undefined it is pretty clearly undefined at \(t=1\) because you have \(\sqrt{t+1}\) in the denominator, and you cannot divide by zero
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
so your next job is to see where the derivative is equal to zero that is, set \[1\frac{2}{\sqrt{t+1}}+0\]and solve for \(t\)
 one year ago

EphiloBest ResponseYou've already chosen the best response.0
so you put zero in place of t right
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
no, you set the derivative equal to zero and solve for \(t\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
i.e. solve \[1\frac{2}{\sqrt{t+1}}+0\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
\[1\frac{2}{\sqrt{t+1}}=0\]
 one year ago

S.RaviTejaBest ResponseYou've already chosen the best response.0
t=+sqrt(2)1,sqrt(2)1
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
@S.RaviTeja no, that is not the solution
 one year ago

S.RaviTejaBest ResponseYou've already chosen the best response.0
now substitute these values in second derivative of the function and if value is greater than zero its minima and less than zero maxima.
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
@Ephilo don't stress over this one \[1\frac{2}{\sqrt{t+1}}=0\iff\sqrt{t+1}=2\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
now you have two critical points, \(t=0\) and \(t=3\)
 one year ago

S.RaviTejaBest ResponseYou've already chosen the best response.0
oh sorry in a hurry id did a mistake t=3 and 6
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
now it is time for the second derivative take the second derivative and evaluate at the critical points
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
@S.RaviTeja 6 is not a solution
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
second derivative is ... ?
 one year ago

EphiloBest ResponseYou've already chosen the best response.0
okay im working on it @satellite73
 one year ago

S.RaviTejaBest ResponseYou've already chosen the best response.0
ok gotcha:) if we take 6 we get imaginary values if we substitute in the original eqn or any other eqn..Thanks for referring:)
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
ok. use the first answer you wrote for the derivative and use the power rule. then rewrite in radical notation so you can actually compute the number
 one year ago

S.RaviTejaBest ResponseYou've already chosen the best response.0
second derivative is (t+1)^(3/2)
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
\[\frac{1}{\sqrt{t+1}^3}\]
 one year ago

S.RaviTejaBest ResponseYou've already chosen the best response.0
yep thats what i was referring to as derivative that is of satellite73
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
replace \(t\) by \(1\) and see that it is undefined so forget that, it is the left point of the domain then replace \(t\) by \(3\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
since the number you get is clearly positive (it is \(\frac{1}{8}\)) this means the function is concave up dw:1364049678258:dw
 one year ago

EphiloBest ResponseYou've already chosen the best response.0
so im evaluating right ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
no i get \(\frac{1}{8}\) but in any case it is positive which is all you really care about
 one year ago

satellite73Best ResponseYou've already chosen the best response.4
that is not all you have to say that since the second derivative evaluated at the critical point is a positive number, the function is concave up, meaning the critical point is a local minimum
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.