If #y = x^2 ln x#, what are the points of inflection, concavity and critical points?
1 Answer
Critical point at
Explanation:
Finding critical points:
For the function
Thus, to find critical values, we must find the derivative of the function. To do this to
#y'=lnxd/dx(x^2)+x^2d/dx(lnx)#
#y'=lnx(2x)+x^2(1/x)#
#y'=2xlnx+x#
The critical points are where
#2xlnx+x=0#
#x(2lnx+1)=0#
This can be split into two equations equalling
#x=0#
This potential critical point is discarded since
#2lnx+1=0#
#lnx=-1/2#
#x=e^(-"1/2")=1/sqrte#
This is the only critical value:
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
#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
Our test points surrounding
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
Since it's positive at
Since the concavity shifts at