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HELP PLEASE?! use the Second Derivative Test to find all relative extrema.

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i guess you need the first two derivatives what did you get for the first one?

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Other answers:

@satellite73 give me a second im gonna do it right now
the first derivate is
yes it is i would write it as \[1-\frac{2}{\sqrt{t+1}}\] but that is correct.
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 \(t\)
a bit confused here?
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
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
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 \(t\)
is that okay?
so you put zero in place of t right
no, you set the derivative equal to zero and solve for \(t\)
i.e. solve \[1-\frac{2}{\sqrt{t+1}}+0\]
damn typo!!
thats better
okay ill give it a try
@S.RaviTeja no, that is not the solution
now substitute these values in second derivative of the function and if value is greater than zero its minima and less than zero maxima.
@Ephilo don't stress over this one \[1-\frac{2}{\sqrt{t+1}}=0\iff\sqrt{t+1}=2\]
so t=3 ?
yay! lol
now you have two critical points, \(t=0\) and \(t=3\)
oh sorry in a hurry id did a mistake t=3 and -6
now it is time for the second derivative take the second derivative and evaluate at the critical points
@S.RaviTeja -6 is not a solution
second derivative is ... ?
okay im working on it @satellite73
ok gotcha:) if we take -6 we get imaginary values if we substitute in the original eqn or any other eqn..Thanks for referring:)
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
second derivative is (t+1)^(-3/2)
would it be
close enough lol
yep thats what i was referring to as derivative that is of satellite73
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 \(3\)
since the number you get is clearly positive (it is \(\frac{1}{8}\)) this means the function is concave up |dw:1364049678258:dw|
so im evaluating right ?
no i get \(\frac{1}{8}\) but in any case it is positive which is all you really care about
its 1/8
i forgot to simplify
would that be all
oh okay lol
thanks guys (:
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
oh okay @satellite73

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