Magnitude of the Second Derivative

[StanEvans]

So to find the x values of the stationary points on the curve:
f(x)=x³+3x²
you make f '(x)=0
so:
3x²+6x=0
x=0 or x=-2
Then to find which of these points are maximum or minimum you do f ''(0) and f ''(-2)
so:
6(0)+6=6
6(-2)+6=-6
so the maximum has an x value of -2 and the minimum has an x value of 0.

My question is does the magnitude of a value from the second derivative have any significance?
and if so what does it show or mean?

 

[HallsofIvy]

The 2nd degree Taylor polynomial for this function, about x= 0, is $y= 0+ 0x+ 6x^2= 6x^2$. That tells you that this is, in the neighborhood of x= 0, a parabola. The fact that "6" is positive tells you that it opens upward (so x= 0 is a minimum) and the absolute value tells you how fast it increases as well as the curvature at that point.

The 2nd degree Taylor polynomial for this function, about x= -2, is $y= 4+ 0(x- 2)- 6(x- 2)^2= 4- 6(x- 2)^2$. That tells you that this is, in the neighborhood of x= 2, a parabola. The fact that "-6" is negative tells you that it opens downward (so x= -2 is a maximum) and the absolute value tells you how fast it decreases as well as the curvature at that point.

 

[StanEvans]

The 2nd degree Taylor polynomial for this function, about x= 0, is $y= 0+ 0x+ 6x^2= 6x^2$. That tells you that this is, in the neighborhood of x= 0, a parabola. The fact that "6" is positive tells you that it opens upward (so x= 0 is a minimum) and the absolute value tells you how fast it increases as well as the curvature at that point.

The 2nd degree Taylor polynomial for this function, about x= -2, is $y= 4+ 0(x- 2)- 6(x- 2)^2= 4- 6(x- 2)^2$. That tells you that this is, in the neighborhood of x= 2, a parabola. The fact that "-6" is negative tells you that it opens downward (so x= -2 is a maximum) and the absolute value tells you how fast it decreases as well as the curvature at that point.

ok I think I understand,
just I am not sure what it is that you mean by a 2nd Taylor polynomial.

 

[Ssnow]

As @HallsofIvy said geometrically the second derivative of a function $f(x)$ is proportional with the curvature of the plane curve represented by $f$ by the formula $\kappa=\frac{|f''|}{(1+f'^2)^{3/2}}$.

 

[Mark44]

My question is does the magnitude of a value from the second derivative have any significance?
and if so what does it show or mean?

The 2nd derivative gives the rate of change of the 1st derivative. If the 2nd derivative is positive, that means that the 1st derivative is increasing, which could mean that it's changing slope from more negative to less negative, or from negative to positive, or from less positive to more positive. You apply this test at critical points where the 1st derivative is 0. So if the slope of the tangent lines to the curve is negative to the left of the critical point, but positive to the right, then the critical point is a relative minimum.

The situation is similar when the 2nd derivative is negative, but here we're dealing with a relative maximum.

 

[Mark44]

My question is does the magnitude of a value from the second derivative have any significance?

The magnitude doesn't have much significance -- just its sign.

 

[StanEvans]

ok thank you for helping