<prompt> Fumana bonyane ba boleng ba tšebetsofx, y 2x2 y3  <response>: Instruction in English: Find the minimum value of the function . Response in English: To find the minimum value of the function fx, y 2x2 y3, we can differentiate it with respect to x and y and set the derivatives to zero to find the critical points of the function.The partial derivative of fx, y with respect to x is fxx, y 4x.The partial derivative of fx, y with respect to y is fyx, y 3y2.Setting fx to 0, we get x 0.Substituting x 0 into fy, we get 0 3y2.But, there is no value of y that satisfies this equation.Alternatively, we can also look at the behavior of the function as x and y tend to positive and negative infinity. As x increases in either direction, 2x2 increases as well, towards positive infinity. As y increases towards positive infinity, so does y3, but as y decreases towards negative infinity, y3 decreases towards negative infinity. This shows that the function doesnt have a minimum value.Thus, the function fx, y 2x2 y3 does not have a minimum value. . Response in Sesotho: Ho fumana bonyane ba boleng ba mosebetsi fx, y 2x2 y3, re ka e khetholla mabapi le x le y le ho beha li-derivatives ho zero ho fumana lintlha tsa bohlokoa tsa mosebetsi. Karolo e ngoe e nkiloeng ho fx, y mabapi le x ke fxx, y 4x. Karolo e ngoe e nkiloeng ho fx, y mabapi le y ke fyx, y 3y2. Ho beha fx ho 0, re fumana x 0. Ho kenya x 0 ho fy, re fumana 0 3y2. Empa, ha ho na boleng ba y bo khotsofatsang palo ena. Ntle le moo, re ka sheba boitsoaro ba tsebetso joalo ka x le y li na le bosafeleng bo botle le bo bobe. Ha x e ntse e eketseha ka lehlakoreng le leng, 2x2 le eona e ea eketseha, ho ea ho infinity e ntle. Ha y e ntse e eketseha ho ea ho positive infinity, le y3 e ntse e fokotseha, empa ha y e fokotseha ho ea ho infinity e mpe, y3 e fokotseha ho ea ho infinity e mpe. Sena se bontša hore tšebetso ha e na boleng bo tlase. Kahoo, tshebetso fx, y 2x2 y3 ha e na boleng bo tlase. 