Mitchtwitchita
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can anybody tell me what the critical point of xe^x is? When I try putting it into my calculator, it just shows a line staring at zero with an asymptote at x=1.
The critical point of the function y = xe^x is located at x = -1, where the derivative is equal to zero. The derivative is calculated as ∂/∂x (xe^x) = xe^x + e^x. The confusion arose from a misinterpretation of the function's behavior at x = 1, which is not an asymptote, as the function has a defined value at that point. Understanding critical points is essential for analyzing the function's local behavior and applying the inverse function theorem.
PREREQUISITESStudents and professionals in mathematics, particularly those studying calculus, as well as educators looking for examples of critical point analysis in functions.
Then you don't got it. If "when x= 1, the answer is -1" is in response to my question "What is the value of y= xex when x= 1?" (my point being that if it has a value, x= 1 cannot be an asyptote), then when x= 1, y= xex= 1(e1)= 1. I can't imagine how you would get a negative number for that. And you have already been told that the critical point is NOT at x= 1.Mitchtwitchita said:Thanks for your help guys. I think I got it now. When x=1, the answer is -1. This is the critical point for the function. I don't what the deal was with my calculator. It could have been the scale. Anyhoo, thanks again.