# 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 **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

#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