Find all critical points of h(t) = (t2 − 36)^1/3 and determine the extreme values of h(t) on
[0, 6] and [0, 12]. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

I have found the critical points to be -6,0,6.
When I do the first derivative test on a number line for [0,6], I get a decreasing from (-INFINITY,0) and increasing from (0,6) U (6, INFINITY). I said that the maximum value DNE and that the minimum value for the interval [0,6] was at h(t)=0. I got this marked wrong, can anyone help me to the right answer?

Question

Find all critical points of h(t) = (t2 − 36)^1/3 and determine the extreme values of h(t) on 
[0, 6] and [0, 12]. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

I have found the critical points to be -6,0,6.
When I do the first derivative test on a number line for [0,6], I get a decreasing from (-INFINITY,0) and increasing from (0,6) U (6, INFINITY). I said that the maximum value DNE and that the minimum value for the interval [0,6] was at h(t)=0. I got this marked wrong, can anyone help me to the right answer?

sshayer · Answer

$$h(t)=\left( t^2-36 \right)^{\frac{ 1 }{ 3 }},h \prime \left( t \right)=\frac{ 2t }{ \left( t^2-36 \right)^{\frac{ 2 }{ 3 }} }$$
$$h'(t)=0~gives~t=0$$
and extreme value at t=0 is
$$h(0)=(-36)^{\frac{ 1 }{ 3 }}$$