2 years ago

PLEASE HELP ME!!

1 answer:

5 0

Take the derivative:
g’(x) = 12x^3 - 24x^2

Set equal to zero and solve:
0 = 12x^3 - 24x^2
0 = 12x^2 (x - 2)

x = 0 or x = 2

Plug back into original
g(0) = 3(0^4) - 8(0^3)
g(0) = 0 - 0
g(0) = 0

g(2) = 3(2^2) - 8(2^3)
g(2) = 3(4) - 8(8)
g(2) = 12 - 64
g(2) = -52

There is an absolute max at (0,0) or when x = 0

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