How do you use the first and second derivatives to sketch #y = x - ln |x|#?

Question

Luke Alvarez · Accepted Answer

#y(x)# is monotone increasing in #(-oo,0)# and #(1,+oo)#, decreasing in #(0,1)#, reaches a local minimum for #x=1# and is concave up in its whole domain.

We can calculate the first derivative separately for #x<0# and #x>0#: 1) For #x<0# we have: #y=x-ln(-x)# #(dy)/(dx) = 1-1/x# 2) For #x>0# we have: #y=x-ln(x)# #(dy)/(dx) = 1-1/x# Thus the derivative is the same in the two intervals #(-oo,0)# and #(0,+oo)#. The second derivative is: #(d^2y)/(dy^2) = d/(dx) (1-1/x)= 1/x^2# We can therefore see that the function has only a critical point for #x=1# and that #y'(x) < 0# for #x in (-oo,0) uu (0,1)# #y'(x) >0# for #x in (1,+oo)# so this critical point is a local minimum. It has no inflection points and is concave up in its domain. We can also note that: #lim_(x->-oo) y(x) = -oo# #lim_(x->+oo) y(x) = +oo# #lim_(x->0) y(x) = +oo# So #y(x)# starts from #-oo# strictly increases approaching #+oo# for #x->0^-#, then decraeses starting from #+oo# as #x->0^+#, reaches a minimum for #x=1# and then starts increasing again approaching #+oo# for #x->+oo#.

Luke Alvarez · Answer

undefined To sketch the graph of $ y = x - \ln |x| $ using the first and second derivatives:

1. Find the first derivative of $ y $ with respect to $ x $.
2. Find critical points by setting the first derivative equal to zero and solving for $ x $.
3. Determine the intervals of increase and decrease by analyzing the sign of the first derivative.
4. Find the second derivative of $ y $ with respect to $ x $.
5. Determine the intervals of concavity by analyzing the sign of the second derivative.
6. Find any inflection points by setting the second derivative equal to zero and solving for $ x $.
7. Sketch the graph using the information obtained from steps 3, 5, and 6, along with any additional points of interest such as intercepts or asymptotes.