The discussion revolves around identifying points on the graph of the function y = x3ex where the tangent line is horizontal. Participants are exploring the application of the product rule in calculus to find the derivative and determine the conditions for horizontal tangents.
The discussion is ongoing, with participants providing feedback on each other's attempts. Some guidance has been offered regarding the identification of points with horizontal tangents, and there is an acknowledgment of the need to consider all factors in the derivative.
Participants note uncertainty about the correctness of their calculations and the implications of the factors in the derivative. There is a focus on ensuring that all potential solutions are considered in the context of the problem.
Dustobusto said:Homework Statement
Find the points on the graph of y = x3ex where the tangent line is horizontal.
(Your answer should include x-values and y-values)Homework Equations
Product rule
fg'+gf'The Attempt at a Solution
(x3)(ex)+(ex)(3x2)
x3ex+3x2ex= y'
x3ex+3x2ex= 0
x2ex(x+3) = 0
x = -3
Going back to original equation, and plugging in x:
y = -33e-3
y = -27e-3
y ≈-1.34
Not sure if I did this correctly.
Mark44 said:Looks fine, but you should provide a summary of what you found - something like this: The only point with a horizontal tangent is (-3, -27e-3), or approximately (-3, -1.34).