What are the critical points of # f(x) = e^xlnx^2#?

Question

Charles Arocha · Accepted Answer

The function #f(x) = e^x ln(x^2)# has a single critical point in #(-oo,0)# whose value is the solution of the equation:

#xln(-x) +1 = 0# 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]}

Emilia Aguilar · Answer

undefined To find the critical points of $ f(x) = e^x \ln(x^2) $, we first need to compute its derivative and then find where the derivative equals zero or is undefined.

Given that $ f(x) = e^x \ln(x^2) $, the derivative of $ f(x) $ with respect to $ x $, denoted as $ f'(x) $, can be found using the product rule of differentiation:

$$ f'(x) = \left(e^x \cdot \frac{d}{dx}\ln(x^2)\right) + \left(\ln(x^2) \cdot \frac{d}{dx}e^x\right) $$

Applying the chain rule and derivative of natural logarithm, we have:

$$ f'(x) = \left(e^x \cdot \frac{2x}{x^2}\right) + \left(\ln(x^2) \cdot e^x\right) $$

Simplify the expression:

$$ f'(x) = \frac{2xe^x}{x} + \ln(x^2)e^x $$

$$ f'(x) = 2e^x + 2x\ln(x)e^x $$

To find critical points, we set $ f'(x) $ equal to zero and solve for $ x $:

$$ 2e^x + 2x\ln(x)e^x = 0 $$

$$ 2e^x(1 + x\ln(x)) = 0 $$

The only solution to this equation is when $ 1 + x\ln(x) = 0 $, since $ e^x $ is never zero. Solving this equation for $ x $ would involve numerical methods or graphical analysis.

Therefore, the critical points of $ f(x) = e^x \ln(x^2) $ occur at the solutions to $ 1 + x\ln(x) = 0 $.