No it is NOT always bounded. Here is an example of an unbounded one. 1. 2x-y>-2 2. 4x+y
Assuming the domain is unbounded, the linear function continues to be a linear function to its end.
What is the area bounded by the graph of the function f(x)=1-e^-x over the interval [-1, 2]?
Surely, you should check the value of the function at the boundaries of the region first. Rest depends on what the function is.
The inverse of the inverse is the original function, so that the product of the two functions is equivalent to the identity function on the appropriate domain. The domain of a function is the range of the inverse function. The range of a function is the domain of the inverse function.
No. y = 1/x is continuous but unbounded.
bounded signal
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Bounded media use physical connectors, such as cables or wires, to transmit signals between devices, while unbounded media use wireless transmission methods, like radio waves or infrared signals. Bounded media have a defined path for signals to travel, while unbounded media allows for more flexibility in signal transmission without physical constraints.
I believe the maximum would be two - one when the independent variable tends toward minus infinity, and one when it tends toward plus infinity. Unbounded functions can have lots of asymptotes; for example the periodic tangent function.
No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.No, a measurable function may have a finite number of discontinuities (for the Riemann measure), or a countably infinite number of discontinuities (for the Lebesgue measure). It should also be bounded (have some upper and lower bound, or limit, in the domain that is being measured), to be measureable. At least, some unbounded functions are not measurable.
No it is NOT always bounded. Here is an example of an unbounded one. 1. 2x-y>-2 2. 4x+y
A set of numbers is bounded if there exist two numbers x and y (with x ≤ y)such that for every member of the set, x ≤ a ≤ y. A set is unbounded if one or both of x and y is infinite. Similar definitions apply for sets in more than 1 dimension.
Assuming the domain is unbounded, the linear function continues to be a linear function to its end.
An unbounded solution refers to a scenario in linear programming where the objective function can be infinitely improved without reaching an optimal solution. This typically occurs when the feasible region is not bounded and extends infinitely in one or more directions. In such cases, the problem is said to have an unbounded solution because the value of the objective function can be made arbitrarily large or small without limit.
NO
The graphical method for solving LPP in two unknowns is as follows: 1)Graph the feasible region. 2)Compute the coordinates of the corner points. 3)Substitute the coordinates of the corner points into the objective function to see which gives the optimal value. 4)If the feasible region is bounded,this method can be misleading:optimal solutions always exist when the feasible region is bounded,but may or may not exist when the feasible region is unbounded.