Natural Logs Tangents

[thelazyman]

Hi, please someone help me with this question.


Find all points at which the tangent to the curve defined by y=x^2e^-x is horizontal.

 

[Unco]

A horizontal line has a slope of zero, so the slope of the tangent to the curve at those points is zero.

The slope of the tangent at a particular point on the curve is given by the derivative.

Determine \(\displaystyle \mbox{\frac{dy}{dx}}\) and set it to zero. Solve for \(\displaystyle \mbox{x}\).

(Galactus illustrated how to solve such an equation just today at http://www.freemathhelp.com/forum/viewtopic.php?t=10764)

That gives the x-ordinates of the points where the tangent to the curve's slope is zero. Use the curve's equation to determine the corresponding y-ordinates.

 

[pka]

Note that e<SUP>x</SUP> is never zero for any x

If we are given that \(\displaystyle \L
y = x^2 e^{ - x}\) then

\(\displaystyle \L
y' = 2xe^{ - x} - x^2 e^{ - x} = xe^{ - x} \left( {2 - x} \right)\).

Where is y’=0?