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

- anonymous

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

- schrodinger

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- anonymous

|dw:1364048045697:dw|

- anonymous

\[h(t)=t-4\sqrt{t+1}\]?

- anonymous

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

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## More answers

- anonymous

@satellite73 give me a second im gonna do it right now

- anonymous

ok

- anonymous

the first derivate is

- anonymous

|dw:1364048369021:dw|

- anonymous

yes it is
i would write it as
\[1-\frac{2}{\sqrt{t+1}}\] but that is correct.

- anonymous

okay

- 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

a bit confused here?

- 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

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

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

is that okay?

- anonymous

so you put zero in place of t right

- anonymous

no, you set the derivative equal to zero and solve for \(t\)

- anonymous

i.e. solve
\[1-\frac{2}{\sqrt{t+1}}+0\]

- anonymous

damn typo!!

- anonymous

\[1-\frac{2}{\sqrt{t+1}}=0\]

- anonymous

t=+sqrt(2)-1,-sqrt(2)-1

- anonymous

thats better

- anonymous

okay ill give it a try

- anonymous

@S.RaviTeja no, that is not the solution

- 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

@Ephilo don't stress over this one
\[1-\frac{2}{\sqrt{t+1}}=0\iff\sqrt{t+1}=2\]

- anonymous

so t=3 ?

- anonymous

yes

- anonymous

yay! lol

- anonymous

now you have two critical points, \(t=0\) and \(t=3\)

- anonymous

oh sorry in a hurry id did a mistake t=3 and -6

- anonymous

now it is time for the second derivative
take the second derivative and evaluate at the critical points

- anonymous

@S.RaviTeja -6 is not a solution

- anonymous

second derivative is ... ?

- anonymous

okay im working on it @satellite73

- 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

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

second derivative is (t+1)^(-3/2)

- anonymous

would it be

- anonymous

|dw:1364049467577:dw|

- anonymous

\[\frac{1}{\sqrt{t+1}^3}\]

- anonymous

close enough lol

- anonymous

yep thats what i was referring to as derivative that is of satellite73

- 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

since the number you get is clearly positive (it is \(\frac{1}{8}\)) this means the function is concave up |dw:1364049678258:dw|

- anonymous

so im evaluating right ?

- anonymous

yes

- anonymous

1/16

- anonymous

no i get \(\frac{1}{8}\) but in any case it is positive which is all you really care about

- anonymous

its 1/8

- anonymous

i forgot to
simplify

- anonymous

would that be all

- anonymous

yep

- anonymous

oh okay lol

- anonymous

thanks guys (:

- 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

oh okay @satellite73

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