# Math Help - Second derivative = 0

1. ## Second derivative = 0

When we're trying to figure out whether a critical point is a maximum or a minimum we can use second derivative test, and if we find that the point of our interest has second derivative equal to zero, this means that there can still be a maximum or a minimum at this point.

Why is this the case?

2. ## Re: Second derivative = 0

Easiest to look at an example. Consider the two functions

$f(x) = x^4, g(x) = - x^4$

Both have critical points at $x = 0$ yet both also have their second derivatives vanish at $x = 0$ so the second derivative test is inconclusive for this case. However, it's not hard to see that

$f \ge 0$ and $g \le 0$

so for $f$, it has a minimum at $x = 0$ whereas $g$ has a maximum there.

3. ## Re: Second derivative = 0

Does someone have another example?

4. ## Re: Second derivative = 0

Funny, so there are two distinct cases:

1. the rate of change of a function is zero at a particular point but it is continuing to grow (functions like $x^2$)
2. the rate of change of a function is zero at a particular point and the function sort of lingers at it, while x is at this value (and in the case of $x^4$ immediately starts increasing just after it gets past this point, where the slope is zer)

5. ## Re: Second derivative = 0

Hello, maxpancho!

When we're trying to figure out whether a critical point is a maximum or a minimum,
we can use second derivative test. .And if we find that the point of our interest has
a second derivative equal to zero, there can still be a maximum or a minimum at this point.

Why is this the case?

Let $x = c$ be the critical point.

If $f''(c) > 0$, the graph is concave up: . $\smile$

If $f''(c) < 0$, the graph is concave down: . $\frown$

If $f''(c) = 0$, the graph is neither concave up nor concave down.
. . Using baby-talk, the graph "flattens" briefly at that point.
The point could be a maximum, a minimum, or an inflection point.
. . We use additional tests to determine which it is.

6. ## Re: Second derivative = 0

So this tells us that a parabola like $x^2$ doesn't flatten at zero, but something like... well, it seems that at least when we consider $x^n$, when $n \in \mathbb{Z}$, $n > 2$ the function flattens at zero. Interesting.