CLP-1 Differential Calculus

Subsection 3.5.1 Local and Global Maxima and Minima

We start by asking:

Suppose that the maximum (or minimum) value of \(f(x)\) is \(f(c)\) then what does that tell us about \(c\text{?}\)

Notice that we have not yet made the ideas of maximum and minimum very precise. For the moment think of maximum as “the biggest value” and minimum as “the smallest value”.

Now back to thinking about what happens around a maximum. Suppose that the maximum value of \(f(x)\) is \(f(c)\text{,}\) then for all “nearby” points, the function should be smaller.

Consider the derivative of \(f'(c)\text{:}\)

Split the above limit into the left and right limits:

• Consider points to the right of \(x=c\text{,}\) For all \(h \gt 0\text{,}\)
  \begin{align*} f(c+h) & \le f(c) & \text{which implies that}\\ f(c+h)-f(c) &\le 0 & \text{which also implies}\\ \frac{f(c+h)-f(c)}{h} & \le 0 & \text{since } \frac{\text{negative}}{\text{positive}} = \text{negative}. \end{align*}
  But now if we squeeze \(h \to 0\) we get
  \begin{align*} \lim_{h \to 0^+} \frac{f(c+h)-f(c)}{h} &\leq 0 \end{align*}
  (provided the limit exists).
• Consider points to the left of \(x=c\text{.}\) For all \(h \lt 0\text{,}\)
  \begin{align*} f(c+h) & \le f(c) & \text{which implies that}\\ f(c+h)-f(c) &\le 0 & \text{which also implies}\\ \frac{f(c+h)-f(c)}{h} & \ge 0 & \text{since } \frac{\text{negative}}{\text{negative}} = \text{positive}. \end{align*}
  But now if we squeeze \(h \to 0\) we get
  \begin{align*} \lim_{h \to 0^-} \frac{f(c+h)-f(c)}{h} &\geq 0 \end{align*}
  (provided the limit exists).
• So if the derivative \(f'(c)\) exists, then the above right- and left-hand limits must agree, which forces \(f'(c) = 0\text{.}\)

Thus we can conclude that

If the maximum value of \(f(x)\) is \(f(c)\) and \(f'(c)\) exists, then \(f'(c)=0\text{.}\)

Using similar reasoning one can also see that

If the minimum value of \(f(x)\) is \(f(c)\) and \(f'(c)\) exists, then \(f'(c)=0\text{.}\)

Notice two things about the above reasoning:

• Firstly, in order for the argument to work we only need that \(f(x) \lt f(c)\) for \(x\) close to \(c\) — it does not matter what happens for \(x\) values far from \(c\text{.}\)
• Secondly, in the above argument we needed to consider \(f(x)\) for \(x\) both to the left of and to the right of \(c\text{.}\) If the function \(f(x)\) is defined on a closed interval \([a,b]\text{,}\) then the above argument only applies when \(a \lt c \lt b\) — not when \(c\) is either of the endpoints \(a\) and \(b\text{.}\)

Consider the function below

This function has only 1 maximum value (the middle green point in the graph) and 1 minimum value (the rightmost blue point), however it has 4 points at which the derivative is zero. In the small intervals around those points where the derivative is zero, we can see that function is locally a maximum or minimum, even if it is not the global maximum or minimum. We clearly need to be more careful distinguishing between these cases.

Consider again the function we showed in the figure above

It has 2 local maxima and 2 local minima. The global maximum occurs at the middle green point (which is also a local maximum), while the global minimum occurs at the rightmost blue point (which is not a local minimum).

Using the above definition we can summarise what we have learned above as the following theorem ¹This is one of several important mathematical contributions made by Pierre de Fermat, a French government lawyer and amateur mathematician, who lived in the first half of the seventeenth century.:

• It is often (but not always) the case that, when \(f(x)\) has a local maximum at \(x=c\text{,}\) the function \(f(x)\) increases strictly as \(x\) approaches \(c\) from the left and decreases strictly as \(x\) leaves \(c\) to the right. That is, \(f'(x) \gt 0\) for \(x\) just to the left of \(c\) and \(f'(x) \lt 0\) for \(x\) just to the right of \(c\text{.}\) Then, it is often the case, because \(f'(x)\) is decreasing as \(x\) increases through \(c\text{,}\) that \(f''(c) \lt 0\text{.}\)
• Conversely, if \(f'(c)=0\) and \(f''(c) \lt 0\text{,}\) then, just to the right of \(c\) \(f'(x)\) must be negative, so that \(f(x)\) is decreasing, and just to the left of \(c\) \(f'(x)\) must be positive, so that \(f(x)\) is increasing. So \(f(x)\) has a local maximum at \(c\text{.}\)
• Similarly, it is often the case that, when \(f(x)\) has a local minimum at \(x=c\text{,}\) \(f'(x) \lt 0\) for \(x\) just to the left of \(c\) and \(f'(x) \gt 0\) for \(x\) just to the right of \(c\) and \(f''(x) \gt 0\text{.}\)
• Conversely, if \(f'(c)=0\) and \(f''(c) \gt 0\text{,}\) then, just to the right of \(c\) \(f'(x)\) must be positive, so that \(f(x)\) is increasing, and, just to the left of \(c\) \(f'(x)\) must be negative, so that \(f(x)\) is decreasing. So \(f(x)\) has a local minimum at \(c\text{.}\)

Theorem 3.5.4 says that, when \(f(x)\) has a local maximum or minimum at \(x=c\text{,}\) there are two possibilities.

• The derivative \(f'(c)=0\text{.}\) This case is illustrated in the following figure.

  

  Observe that, in this example, \(f'(x)\) changes continuously from negative to positive at the local minimum, taking the value zero at the local minimum (the red dot).

• The derivative \(f'(c)\) does not exist. This case is illustrated in the following figure.

  

  Observe that, in this example, \(f'(x)\) changes discontinuously from negative to positive at the local minimum (\(x=0\)) and \(f'(0)\) does not exist.

This theorem demonstrates that the points at which the derivative is zero or does not exist are very important. It simplifies the discussion that follows if we give these points names.

We'll now look at a few simple examples involving local maxima and minima, critical points and singular points. Then we will move on to global maxima and minima.