Question

How do you use the second derivative test to find the relative extrema for `h(t) = t-4sqrt(t+1)`?

Answer 1

See the explanation.

Explanation:

`t+1>=0 <=> t>=-1`

`D_f=[-1,+oo)`

`h'(t)=1-2/sqrt(t+1)`
`h''(t)=1/((t+1)sqrt(t+1))`

`h'(t)=0 <=> 2/sqrt(t+1)=1 <=> sqrt(t+1)=2 <=>`

`<=> t+1=4 ^^ t+1>=0 <=> t=3`

`h''(3)=1/(4sqrt4)=1/8>0, h(t)` has a minimum for `t=3`.