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) wheny''>0 .y has a point of inflection wheny''=0 and the concavity shiftsy is concave downwards (concave) wheny''<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