Find all x-value(s) where the first derivative of g(x)= (e^x )*x^2 has a horizontal tangent line?

need to find the derivative of g′(x).

1 Answer
Feb 16, 2017

x=-2+-sqrt2
[x~=-0.5858 or x~=-3.4142]

Explanation:

g(x) = e^x*x^2

Applying the Product Rule

g'(x)= e^x*2x+e^x*x^2

=e^x(2x+x^2) = e^x*x(x+2)

And: g''(x) = e^x(2+2x) + e^x(2x+x^2)

=e^x(x^2+4x+2)

A tangent to g'(x) will be horizontal where g''(x)=0

That is, where: e^x(x^2+4x+2)=0

Since e^x>0 forall x -> x^2+4x+2=0

Applying the quadratic formula:

x=(-4+-sqrt(16-4*1*2))/2 = (-4+-sqrt8)/2

=-2+-sqrt2

[x~=-0.5858 or x~=-3.4142]

This result can be seen by the graph of g'(x) below:

graph{e^x*(x^2+2x) [-5.733, 2.063, -1.654, 2.242]}