Simplify the expression noting that based on properties of logarithms:
f(x) = e^x ln(x^2) = 2e^xlnabsx
the critical points are by definition the points where f'(x) = 0 so evaluate the derivative of the function:
(df)/dx = d/dx (2e^xlnabsx) =2 ((d/dx e^x)lnabsx +e^x (d/dx lnabsx))
(df)/dx =2 (e^xlnabsx +e^x/x)
(df)/dx =2 e^x(lnabsx +1/x)
To identify the critical points then we have to solve the equation:
2 e^x(lnabsx +1/x) = 0
and as 2e^x != 0:
lnabsx=-1/x
This is a logarithmic equation that can be solved only approximately:
note that for x>0 we have lnx = -1/x < 0 so the solution should be in the interval (0,1).
However given:
lnx+1/x = 0
as x !=0 this is equivalent to:
xlnx + 1 = 0
Now consider the function g(x) = xlnx+1:
lim_(x->0) (xlnx+1) = 1+ lim_(x->0) xlnx
lim_(x->0) (xlnx+1) = 1+ lim_(x->0) lnx/(1/x)
Apply l'Hospital's:
lim_(x->0) (xlnx+1) = 1+ lim_(x->0) (d/dx lnx)/(d/dx 1/x)
lim_(x->0) (xlnx+1) = 1+ lim_(x->0) (1/x)/(-1/x^2)= 1+ lim_(x->0) (-x) = 1
while:
lim_(x->1) (xlnx+1) = 1
Consider now:
d/dx (xlnx+1) = 1+lnx
so the only critical point for this function is in x=1/e where its value is:
g(1/e) = 1/eln(1/e)+1 = 1-1/e > 0
and as g''(1/e) = e >0 this is a local minimum. As the minimum value of the function is positive, this means that it is never zero. So actually we have no critical points for x > 0
On the other hand for x < 0 the equation becomes:
xln(-x) +1 =0
and substituting t=-x we can see that:
lim_(x->0^-) (xln(-x)+1) = 1- lim_(t->0^+) tlnt =1
while:
lim_(x->-oo) (xln(-x)+1) = -oo
thus the function assumes positive and negative values and as it is continuous in the interval (-oo,0) it must vanish at least in one point.
graph{e^x ln(x^2) [-10, 10, -5, 5]}
