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
It depends on if the ocean floor is cold or warm or freezing maybe below 0°
Answer:
y-5=12(x+8)
Step-by-step explanation:
G and c you see look 2 times and look at the lines
-18 and +2
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