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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.
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.
Critical points are points on a function where the derivative is equal to zero or does not exist. They are important in calculus because they can help determine the maxima, minima, and inflection points of a function.
To find the critical points of xe^x, we need to take the derivative of the function, which is e^x + xe^x. Then, we set the derivative equal to zero and solve for x. This gives us the critical point x = -1.
In xe^x, the critical point at x = -1 is the minimum point of the function. This means that at this point, the function has the lowest possible value. It is also the point where the function changes from decreasing to increasing.
The critical point at x = -1 affects the graph of xe^x by creating a "turning point" on the graph. Before the critical point, the graph is decreasing, and after the critical point, the graph is increasing. The graph also has a concave up shape at the critical point.
No, there can be other points where the derivative is equal to zero, but not all of them are critical points. Some may be inflection points or points where the function has a horizontal tangent line. Critical points are specifically points where the function changes from increasing to decreasing or vice versa.