What Are the Max and Min Values of the Function f(x) = 1/3X^3 - 1/2^2 - 6x + 4?

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The function f(x) = 1/3X^3 - 1/2^2 - 6x + 4 is unbounded, meaning it does not have a global maximum or minimum. While the initial assertion suggested that the maximum is positive infinity and the minimum is negative infinity, this terminology is misleading. Instead, it is more accurate to state that the function lacks global extrema, as it continues to increase and decrease without bound.

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consider function f(x) = 1/3X^3 - 1/2^2 - 6x + 4

maximum is positive infinity and minimum is negative infiniti correct?

Thank you!
 
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Not if you are looking for local optima. It might be better to say there is no global maximum or minimum, than it is at infinity, to avoid unnecessary use of infinity.
 
Originally posted by physicszman
consider function f(x) = 1/3X^3 - 1/2^2 - 6x + 4

maximum is positive infinity and minimum is negative infiniti correct?

Thank you!

I wouldn't put it that way. I would say that the function is "unbounded" and has no global maximum or minimum.
 
thanks for the help!
 

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