Understanding meaning of $f''$ for $x^2$ and $x^4$

Question

If $y=x^4$

Then $y' = 4x^3$

and $y'' = 12x^2$

At $x=0$, $y'=0$ and $y''=0$.

So, at $x=0$, the gradient is zero is not increasing or decreasing at that point. I can believe this if I look at a plot of $y=x^4$.

Now, $y=x^2$ has a similar shape, so we could draw the same conclusion, right?

$y=x^2$

$y'= 2x$

$y'' = 2$

The gradient is 0 at $x=0$ but since $y''>0$ we must conclude that it is also increasing at this point. Again, I could believe this if I were to look at the graph.

I don't understand why we don't get the same result from the second derivatives of both of these functions.

At the origin, I would expect both graphs to behave the same. They look the same, so why don't we get the same result from their second derivatives?

Answer 1

At the origin, I would expect both graphs to behave the same. They look the same, so why don't we get the same result from their second derivatives?

The graphs only superficially look the same. Here is a look at $y=x^2$ and $y=x^4$ close to the origin:

As you can see, the gradient of $y=x^2$ is steadily increasing at the same rate for all $x$. But $y=x^4$ is so flat that not only is the gradient $0$ at $x=0$, but it does not appear to be changing at all near $x=0$.

Answer 2

Here's a physical model of how higher derivatives work.

Suppose you're driving a car, and you stop for a red light. The car has an automatic transmission, and the shift lever is at "Drive". If you have driven such a car, you know that you need to keep the brake pedal depressed in order to keep it stopped. If your foot were not on the brake pedal, the transmission would transmit some amount of force from the engine to the wheels of the car, accelerating the car forward.

If the brakes were released instantly, then--that is, if we went from enough braking force to keep the car motionless to no braking force at all, with no in-between settings of the brakes--a plot of the car's progress along the road might look much like a parabola, at least for a few moments: the distance moved would be determined by a speed (i.e. gradient of distance) that is initially zero but that increases linearly with every passing millisecond. The sudden change in acceleration could also cause you a bit of discomfort as the back of your seat suddenly pushed against you; you might feel your neck snap back a bit.

But brake pedals don't instantly go from all the way down to all the way up. What actually happens is you start letting up the brake pedal at some rate. At some point the pedal will be down just far enough to cause just enough braking force to hold the car in place; any further raising of the pedal will reduce the braking force enough to allow the transmission to start moving the car. But since the braking force is (for a moment) almost enough to hold the car in place, the car does not accelerate as fast as it would without any braking. Instead, the acceleration of the car goes from zero gradually up to whatever acceleration the engine can provide at idle. (I'm assuming you haven't started stepping on the accelerator yet.)

If we set our time axis so that $t = 0$ at the exact instant when the braking force was just sufficient to stop the car from moving, we see that the acceleration of the car is zero at $t = 0$, but positive at any instant after $t = 0$; and the distance the car has moved is also positive at any instant after $t = 0$.

That's an example of a non-zero third derivative that allows us to start moving a the same instant that we have a zero second derivative.

Now imagine that while waiting at the stop light, you were holding the brake pedal down just barely enough to keep the car stopped. When you release the brake pedal, it does not go from "not moving" to "rising at $10$ cm/sec" (or whatever rate a brake pedal rises at) instantaneously. Rather, when you release the force with which you held the brake pedal down, it accelerates upward. So at $t = 0$, the rate at which the braking force decreases, which gave us the third derivative of the car's distance traveled, is zero, but at any time after $t = 0$, the car has moved. So at $t = 0$, we have zero distance traveled, zero speed, zero second derivative, and zero third derivative, but for any $t > 0$ all those things are positive.

If you take into account that relaxing your leg muscles (to release the brake) is also a gradual process, that makes the fourth derivative zero as well, though the fifth or sixth derivative might be non-zero. And so forth.

In other words, you can think of the motion of a car, or the rise and/or fall of the graph of a function, and something controlled by a possibly quite long chain of causation. You can have several derivatives that are all zero at a particular instant of time or value of $x$, but as long as the next higher derivative is positive, you will see positive motion. With the first three derivatives all zero, a positive fourth derivative causes the third derivative to increase, which starts to cause the second derivative to increase, which starts to cause the first derivative to increase, which starts to cause the function value to increase.

Answer 3

Another perspective: As Théophile points out, the curves only look superficially the same. Informally, $y = x^4$ looks "flatter" at the origin than does $y = x^2$.

Somewhat more formally, the instantaneous slope (the first derivative) of $y = x^2$ is continually increasing at a rate of $2$ per unit of $x$ all the way from left to right. That is what the second derivative being $2$ means, in fact.

In the case of $y = x^4$, that slope does not change, to first order, in the vicinity of the origin, like a vertically ejected projectile momentarily coming to a standstill at the top of its ascent. That is what the second derivative being $0$ means in this case.