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anonymous
 3 years ago
HELP PLEASE?!
use the Second Derivative Test to find all
relative extrema.
anonymous
 3 years ago
HELP PLEASE?! use the Second Derivative Test to find all relative extrema.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364048045697:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[h(t)=t4\sqrt{t+1}\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i guess you need the first two derivatives what did you get for the first one?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 give me a second im gonna do it right now

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the first derivate is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364048369021:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes it is i would write it as \[1\frac{2}{\sqrt{t+1}}\] but that is correct.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 \(t\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0critical 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 \(t\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you put zero in place of t right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, you set the derivative equal to zero and solve for \(t\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i.e. solve \[1\frac{2}{\sqrt{t+1}}+0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[1\frac{2}{\sqrt{t+1}}=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0t=+sqrt(2)1,sqrt(2)1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay ill give it a try

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@S.RaviTeja no, that is not the solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now substitute these values in second derivative of the function and if value is greater than zero its minima and less than zero maxima.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now you have two critical points, \(t=0\) and \(t=3\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh sorry in a hurry id did a mistake t=3 and 6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now it is time for the second derivative take the second derivative and evaluate at the critical points

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@S.RaviTeja 6 is not a solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0second derivative is ... ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay im working on it @satellite73

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok. 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0second derivative is (t+1)^(3/2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364049467577:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{\sqrt{t+1}^3}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep thats what i was referring to as derivative that is of satellite73

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0replace \(t\) by \(1\) and see that it is undefined so forget that, it is the left point of the domain then replace \(t\) by \(3\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since the number you get is clearly positive (it is \(\frac{1}{8}\)) this means the function is concave up dw:1364049678258:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so im evaluating right ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no i get \(\frac{1}{8}\) but in any case it is positive which is all you really care about

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that 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
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