Identifying Horizontal Tangents on the Graph of y = x^3e^x

[Dustobusto]

Homework Statement

Find the points on the graph of y = x³e^x 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)(e^x)+(e^x)(3x²)

x³e^x+3x²e^x= y'

x³e^x+3x²e^x= 0

x²e^x(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]

Homework Statement

Find the points on the graph of y = x³e^x 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)(e^x)+(e^x)(3x²)

x³e^x+3x²e^x= y'

x³e^x+3x²e^x= 0

x²e^x(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.

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).

 

[Dustobusto]

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).

Okay. Also, with

x²e^x(x+3) = 0

x = -3 but does it also = 0? Graphing it, it looks like I could include that value as well.

 

[Mark44]

Sorry, I overlooked that factor of x², so yes, your function has a horizontal tangent at (0, 0).