Second derivative test for a point at which first derivative is zero A very long time ago at high school (specifically A level) I was taught to find maxima and minima in functions $y(x)$ by looking for values of $x$ at which $\frac{dy}{dx}=0$. I could determine whether such a point was a maximum, minimum or point of inflection by seeing whether $\frac{d^2y}{dx^2}$ was negative, positive or zero at the point.
I'm wondering what the caveats are on using this rule. I ask because I was considering the function $y(x) = x^4$ at $x=0$. It seems obvious that there is a minimum at $x=0$, yet $\frac{d^2y}{dx^2}$ is zero, rather than a positive number, at $x=0$.
 A: Observe that $x = 0$ is the relative (local) and absolute (global) minimum of $f(x) = x^4$ since $f(x) = x^4 \geq 0$ for each real number $x$, with equality holding if and only if $x = 0$.
You are asking about the Second Derivative Test.
It states that if $f$ is a twice-differentiable function, $f'(c) = 0$, and
(a) $f''(c) < 0$, then $f(x)$ has a relative (local) maximum at $x = c$;
(b) $f''(c) > 0$, then $f(x)$ has a relative (local) minimum at $x = c$;
(c) $f''(c) = 0$, then the test is inconclusive.
In your example, $f(x) = x^4$, $f'(x) = 4x^3 = 0$ when $x = 0$ and $f''(x) = 12x^2 = 0$ when $x = 0$.  Therefore, the test is inconclusive.
However, finding the relative (local) extrema of $f(x) = x^4$ is easily handled with the First Derivative Test, which states that if $f$ is continuous on a closed interval $[a, b]$ and the derivative exists everywhere in the open interval $(a, b)$, except possibly at point $c$, then
(a) if $f'(x) > 0$ for all $x < c$ and $f'(x) < 0$ for all $x > c$, then $f$ has a relative maximum at $c$;
(b) if $f'(x) < 0$ for all $x < c$ and $f'(x) > 0$ for all $x < c$, then $f$ has a relative minimum at $c$.
What this theorem conveys is that if a continuous function which is differentiable everywhere in an interval except possibly at point $c$ is increasing to the left of $c$ and decreasing to the right of $c$, then it has a relative maximum at $x = c$, while if the function is decreasing to the left of $c$ and increasing to the right of $c$, then it has a relative minimum at $x = c$.
In our case, $f(x) = x^4$ is differentiable at every real number, $f'(x) = 4x^3 < 0$ if $x < 0$, and $f'(x) = 4x^3 > 0$ if $x > 0$, so $f(x) = x^4$ has a relative minimum at $x = 0$ with relative minimum value $f(0) = 0$.
If neither the First Derivative Test nor the Second Derivative Test can be applied, there is a Higher Order Derivative Test.  It states that if $f$ is a real-valued function differentiable a sufficient number of times in an interval $I$, $c \in I$, $f'(c)= f''(c) = f^{(3)}(c) = \cdots = f^{(n)}(c)$, and $f^{(n + 1)}(c) \neq 0$, then
(a)  if $n$ is odd and $f^{( n + 1 )}(c) < 0$, then $c$ is a relative (local) maximum;
(b)  if $n$ is odd and $f^{( n + 1 )}(c) > 0$, then c is a relative (local) minimum;
(c)  if $n$ is even and $f^{( n + 1 )}(c) < 0$, then $c$ is a strictly decreasing point of inflection;
(d)   if $n$ is even and $f^{( n + 1 )}(c) > 0$, then $c$ is a strictly increasing point of inflection.
For the function $f(x) = x^4$, $f'(0) = f''(0) = f^{(3)}(0) = 0$, while $f^{(4)} = 24 > 0$. Hence, the second condition applies (since $n = 3$ is odd), so $x = 0$ is a relative minimum.
A: $\frac{d^2y}{dx^2} > 0$ is not equivalent to the point being a local minimum.
$$ \frac{d^2y}{dx^2} > 0 \Rightarrow \text{The point is a local minimum. (True)}$$
$$ \text{The point is a local minimum.} \Rightarrow \frac{d^2y}{dx^2} > 0 \text{ (False)} $$
$$ \text{The point is a local minimum.} \Rightarrow \frac{d^2y}{dx^2} \ge 0 \text{ (True)} $$
In case of $y=x^4$ at $x=0$, use the definition of a local minimum.
Point $(a, f(a))$ is called a local minimum of $y=f(x)$ if and only if $a$ satisfies all the following conditions:
$\bullet f'(a)=0$.
$\bullet$ There exists an open interval $(\alpha, \beta)$ such that $\alpha < a < \beta$ and $f(a) \le f(x)$ for any $x \in (\alpha, \beta)$.
With this, we can easily obtain that $(0,0)$ is a local minimum of $y=x^4$.
A: For a twice differentiable function we have (WLOG at $0$)
$$f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}2+o(x^2)$$ meaning that for small increments, the function behaves parabolically.
If the point is stationary only the quadratic term remains and
$$f(x)=f(0)+f''(0)\frac{x^2}2+o(x^2)$$ so that you can tell a maximum or a minimum from the sign of $f''$. But if $f''(0)=0$, you have insufficient information on the local behavior, and you have to look at third or higher derivatives.
In your case,
$$x^4=0+0x+0\frac{x^2}2+0\frac{x^3}6+24\frac{x^4}{24}+0.$$
As the only significant term is an even power of $x$ with a positive coefficient, you must have a minimum.
An extremum point with a zero curvature (second derivative) is called a flat point.
