anonymous
  • anonymous
HELP PLEASE?! use the Second Derivative Test to find all relative extrema.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1364048045697:dw|
anonymous
  • anonymous
\[h(t)=t-4\sqrt{t+1}\]?
anonymous
  • anonymous
i guess you need the first two derivatives what did you get for the first one?

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anonymous
  • anonymous
@satellite73 give me a second im gonna do it right now
anonymous
  • anonymous
ok
anonymous
  • anonymous
the first derivate is
anonymous
  • anonymous
|dw:1364048369021:dw|
anonymous
  • anonymous
yes it is i would write it as \[1-\frac{2}{\sqrt{t+1}}\] but that is correct.
anonymous
  • anonymous
okay
anonymous
  • anonymous
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 \(t\)
anonymous
  • anonymous
a bit confused here?
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \(t\)
anonymous
  • anonymous
is that okay?
anonymous
  • anonymous
so you put zero in place of t right
anonymous
  • anonymous
no, you set the derivative equal to zero and solve for \(t\)
anonymous
  • anonymous
i.e. solve \[1-\frac{2}{\sqrt{t+1}}+0\]
anonymous
  • anonymous
damn typo!!
anonymous
  • anonymous
\[1-\frac{2}{\sqrt{t+1}}=0\]
anonymous
  • anonymous
t=+sqrt(2)-1,-sqrt(2)-1
anonymous
  • anonymous
thats better
anonymous
  • anonymous
okay ill give it a try
anonymous
  • anonymous
@S.RaviTeja no, that is not the solution
anonymous
  • anonymous
now substitute these values in second derivative of the function and if value is greater than zero its minima and less than zero maxima.
anonymous
  • anonymous
@Ephilo don't stress over this one \[1-\frac{2}{\sqrt{t+1}}=0\iff\sqrt{t+1}=2\]
anonymous
  • anonymous
so t=3 ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
yay! lol
anonymous
  • anonymous
now you have two critical points, \(t=0\) and \(t=3\)
anonymous
  • anonymous
oh sorry in a hurry id did a mistake t=3 and -6
anonymous
  • anonymous
now it is time for the second derivative take the second derivative and evaluate at the critical points
anonymous
  • anonymous
@S.RaviTeja -6 is not a solution
anonymous
  • anonymous
second derivative is ... ?
anonymous
  • anonymous
okay im working on it @satellite73
anonymous
  • anonymous
ok gotcha:) if we take -6 we get imaginary values if we substitute in the original eqn or any other eqn..Thanks for referring:)
anonymous
  • anonymous
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
anonymous
  • anonymous
second derivative is (t+1)^(-3/2)
anonymous
  • anonymous
would it be
anonymous
  • anonymous
|dw:1364049467577:dw|
anonymous
  • anonymous
\[\frac{1}{\sqrt{t+1}^3}\]
anonymous
  • anonymous
close enough lol
anonymous
  • anonymous
yep thats what i was referring to as derivative that is of satellite73
anonymous
  • anonymous
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 \(3\)
anonymous
  • anonymous
since the number you get is clearly positive (it is \(\frac{1}{8}\)) this means the function is concave up |dw:1364049678258:dw|
anonymous
  • anonymous
so im evaluating right ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
1/16
anonymous
  • anonymous
no i get \(\frac{1}{8}\) but in any case it is positive which is all you really care about
anonymous
  • anonymous
its 1/8
anonymous
  • anonymous
i forgot to simplify
anonymous
  • anonymous
would that be all
anonymous
  • anonymous
yep
anonymous
  • anonymous
oh okay lol
anonymous
  • anonymous
thanks guys (:
anonymous
  • anonymous
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
anonymous
  • anonymous
oh okay @satellite73

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