The discussion focuses on identifying points on the graph of the function y = x3ex where the tangent line is horizontal. The calculations reveal two critical points: (-3, -27e-3), approximately (-3, -1.34), and (0, 0). The horizontal tangents are determined using the product rule for differentiation, leading to the equation x2ex(x + 3) = 0. This confirms that both x = -3 and x = 0 yield horizontal tangents.
PREREQUISITESStudents studying calculus, particularly those focusing on differentiation and tangent lines, as well as educators looking for examples of applying the product rule in real-world scenarios.
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).