Find the absolute extrema of the function $h ( t ) = ( t - 3 ) ^ { 2 / 3 }$ on the closed interval $[ - 1,6 ]$ . Round your answer to two decimal places.
A)The maximum of the function is 1 and the minimum of the function is 0.
B)The maximum of the function is 2.52 and the minimum of the function is 1.
C)The maximum of the function is 2.52 and the minimum of the function is 0.
D)The maximum of the function is1 and the minimum of the function is 2.08.
E)The maximum of the function is 0 and the minimum of the function is 2.08.

Question

Quizplus · Accepted Answer

To find the absolute extrema of $h(t) = (t - 3)^{2/3}$ on the interval $[-1,6]$, we evaluate the function at the endpoints and any critical points within the interval. The critical points occur where the derivative is zero or undefined. The derivative of $h(t)$ is $h'(t) = \frac{2}{3}(t - 3)^{-1/3}$, which is undefined at $t = 3$, indicating a critical point. Evaluating $h(t)$ at $t = -1$, $t = 3$, and $t = 6$ gives $h(-1) = 2.52$, $h(3) = 0$, and $h(6) = 2.08$. Therefore, the maximum value is $2.52$ (at $t = -1$) and the minimum value is $0$ (at $t = 3$).

Find the Absolute Extrema of the Function h(t)=(t−3)2/3h ( t ) = ( t - 3 ) ^ { 2 / 3 }h(t)=(t−3)2/3

Find the Absolute Extrema of the Function $h ( t ) = ( t - 3 ) ^ { 2 / 3 }$