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

1 Answer
Feb 19, 2016

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.