Second derivative test for a point at which first derivative is zero

calculus

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 $x=0$.

Best Answer

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.