Question

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

Answer 1

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.