Question

Use the second derivative test to find all relative extrema. ..... h(t) = t-4 times the root of (t+1).

Answer 1

which one is h(t)?
\(\large h(t)=t-4\sqrt{t+1} \) or \(\large h(t)=(t-4)\sqrt{t+1} \)

Answer 2

The first one.

Answer 3

did you find h'(t) ?

Answer 4

d'(t) = 1 - 2 / root of t+1

Answer 5

write it as t - 4(t + 1)^0.5
h'(t) = 1 - 2(t + 1) ^(-0.5)
equate this to 0 to find turning points on the graph

Answer 6

i made a right mess of that - must be tired
ill try again
0 = 1 - 2 / sqrt(t+1)
sqrt(t+1) = 2
t+ 1 = 4
t = 3 = thats it

Answer 7

now you need to know if thats a minimum or maximum
second derivate h"(t) = 1 / ( t+ 1)^ 1.5

plug t = 3 into this gives a positive value

therefore h(t) has a minimum value at t = 3
and that value is found by plugging t = 3 into t - 4 sort(t + 1) = 3 - 4 sort(4)
= -5

so u have a minimum at (3,-5)