g ' (x) = (2/3)(x^2 - 4)^(-1/3) ( 2x)
= 0 for a max/min
so 2x = 0
x = 0
when x = 0, g(0) = (-4)^(2/3)
or - cuberoot(16)
value at endpoints:
g(-5) = (21)^(2/3)
g(3) = 5^(2/3)
I am sure you can make use of the above to input your answers
Find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. (Round your answers to two decimal places. If an answer does not exist, enter DNE.)
g(x) = (x^2 − 4)^2/3, [−5, 3]
find
absolute maximum (x, y) =
absolute minimum (x, y) = (smaller x-value)
(x, y) = (larger x-value)
1 answer