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If #y = x^2 ln x#, what are the points of inflection, concavity and critical points?

1 Answer
Feb 19, 2016

Answer:

Critical point at #x=1/sqrte#, concave down on #(0,1/e^("3/2"))#, concave up on #(1/e^("3/2"),+oo)#, point of inflection at #x=1/e^("3/2")#

Explanation:

Finding critical points:

For the function #f(x)#, a critical point at #x=c# where #f(c)# exists is a point where either #f'(c)=0# or #f'(c)# doesn't exist.

Thus, to find critical values, we must find the derivative of the function. To do this to #y=x^2lnx#, we must use the product rule.

#y'=lnxd/dx(x^2)+x^2d/dx(lnx)#

#y'=lnx(2x)+x^2(1/x)#

#y'=2xlnx+x#

The critical points are where #y'=0# or where #y'# is undefined where #y# wasn't.

#2xlnx+x=0#

#x(2lnx+1)=0#

This can be split into two equations equalling #0#:

#x=0#

This potential critical point is discarded since #y'# doesn't exist at #x=0#.

#2lnx+1=0#

#lnx=-1/2#

#x=e^(-"1/2")=1/sqrte#

This is the only critical value: #color(red)(x=1/sqrte)#

Finding concavity and points of inflection:

Concavity, convexity, and points of inflection are all dictated by a function's second derivative.

  • #y# is concave upwards (convex) when #y''>0#.
  • #y# has a point of inflection when #y''=0# and the concavity shifts
  • #y# is concave downwards (concave) when #y''<0#.

To find the function's second derivative, again use the power rule.

#y'=2xlnx+x#

#y''=lnxd/dx(2x)+2xd/dx(lnx)+1#

#y''=lnx(2)+2x(1/x)+1#

#y''=2lnx+2+1#

#y''=2lnx+3#

To determine when this is #<0,>0,# or #=0#, first find when it equals just #0#. This could be a spot where the concavity shifts (the sign changes from positive to negative, or vice versa). If the concavity shifts, there is a point of inflection at that point.

#2lnx+3=0#

#lnx=-3/2#

#x=e^(-"3/2")#

#x=1/e^("3/2")#

This is a possible point of inflection. We should test the concavity (sign of the second derivative) around the point #x=1/e^("3/2")#.

Our test points surrounding #x=1/e^("3/2")# can be #x=0.1# and #x=1#, since #1/e^("3/2")approx0.2231#.

At #x=0.1:#

#y''=2ln(0.1)+3approx-1.605#

At #x=1:#

#y''=2ln(1)+3=3#

Since the second derivative is negative at #x=0.1#, we know that the entire interval #(0,1/e^("3/2"))# is concave down.

Since it's positive at #x=1#, we know the entire interval #(1/e^("3/2"),+oo)# is concave up.

Since the concavity shifts at #x=1/e^("3/2")#, we know there is a point of inflection there.