use the second derivative test?!?!?

Question

use the second derivative test to find the relative extrema.
h(t)= t-4sqrt t+1

Answer 1

h(t) = t - 4√t + 1
h'(t) = 1 - 4(1/2)t^(-1/2) = 1 - 2/√t
= 0 <=> √t = 2 <=> t = 4.
h''(t) = -2 (-1/2) t^(-3/2) = t^(-3/2)
h''(4) = 4^(-3/2) = 1/8 > 0, so h has a minimum at t = 4.

Answer 2

The second derivative test is used to find Concavity and points of inflection. But the first derivative test will tell u the relative extrema "if f'(x) changes from + to - or - to + at c then x=c is a relative extrema" first u have to calculate the first derivative. f'(x)=6x-3x^2 then find the critical points by setting F'(x)=0 which gives you 0 and 3 then you make a table interval (*-infinity,0) (0,3) (3,*infinity) test -2 1 4 sign f'(-2)= -24 f(2)= 3 f'(4)=-24 conclusion decreasing increasing decreasing Then you have to write a statement Since the slope of f(x) changes from decreasing to incr. at 0 then x=0 is a relative min " increasing to decr. at 3 then x=3 is a relative max" to find concavity you just repeat the table with the second derivative and if the conclusion is increasing f(x) has upward concavity " decreasing downward concavity

Answer 3

h''(t) = (-1/4) x (-3/4) x (t+1)^(-7/4)