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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 $x=0$.

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  • $\begingroup$ I’d suggest looking at the wiki page. You’ll note that your example is an example of an inconclusive 2nd derivative test. Fortunately, in this specific case, your 2nd derivative is strictly positive arbitrarily close on both sides, so it is indeed a minimum. $\endgroup$
    – adfriedman
    Jan 24 at 12:26
  • $\begingroup$ Thank you for this reference; as I suspected, what I'd been taught was oversimplified. I'd noted that the second derivative was positive arbitrarily close to $x=0$, and I'm glad that you too consider this to be relevant. $\endgroup$ Jan 24 at 12:38
  • $\begingroup$ math.stackexchange.com/questions/4355099/… $\endgroup$
    – dfnu
    Jan 24 at 12:53

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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.

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    $\begingroup$ Comprehensible and comprehensive. Many thanks. I think you're missing "= 0" on the third line of your para starting "There is a higher order [...]". $\endgroup$ Jan 24 at 13:03
  • $\begingroup$ After turning off the computer, I realized that I should have discussed the First Derivative Test, so I have added a discussion of the First Derivative Test and its application to this problem. $\endgroup$ Jan 24 at 19:03
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$\frac{d^2y}{dx^2} > 0$ 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 $y=x^4$.

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  • $\begingroup$ I liked your summary of true and false statements. After that, you seem to be saying: abandon calculus and use arithmetic. Forgive me, I'm just a physicist. $\endgroup$ Jan 24 at 13:11
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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.

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