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

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Answer 1 · 2015-09-21T15:41:11

See the explanation.

$t + 1 \geq 0 \Leftrightarrow t \geq - 1$ $t+1>=0 <=> t>=-1$

$D_{f} = [- 1, + \infty)$ $D_f=[-1,+oo)$

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

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

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

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

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