Find all relative extrema of the function. Use the Second-Derivative
Test when applicable. (If an answer does not exist, enter DNE.)
f'(x) = 30x - 3x^2 = 0
10x - x^2 = 0
x^2 - 10x = 0
x(x - 10) = =
x = 0
x = 10
Now use second derivative test to determine maxima or minima
f"(x) = 30 - 6x
f"(0) = 30 - 6(0) = 30 which is POSITIVE
Therefore, at x = 0 there is a local MINIMUM at the point (0,0)
f"(10) = 30 - 6(10) = -30 which is NEGATIVE
Therefore, at x = 10 there is a local MAXIMUM at the point (10, 500)
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