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

Question

need to find the derivative of g′(x).

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Answer 1 · 2017-02-16T01:31:14

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

#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]}