How do you use the second derivative test to find the relative extrema for #h(t) = t-4sqrt(t+1)#? Calculus Graphing with the Second Derivative First Derivative Test vs Second Derivative Test for Local Extrema 1 Answer Sasha P. Sep 21, 2015 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#. Answer link Related questions What is the first derivative test to determine local extrema? How does the first derivative test differ from the second derivative test? How does the first derivative test work? What is the first derivative test for critical points? What is the first derivative test for local extreme values? What does the first derivative test tell you? How do you know when to use the first derivative test and the second derivative test? How do you use the second derivative test to find the local extrema for #f(x)=sin(x)#? How do you use the second derivative test to find the local extrema for #f(x)=e^(x^2)#? How do you use the second derivative test how do you find the local maxima and minima of #f(x) =... See all questions in First Derivative Test vs Second Derivative Test for Local Extrema Impact of this question 2025 views around the world You can reuse this answer Creative Commons License