Maxima and minima treated analytically

By making use of the results of the last two sections we can now give a general discussion of maxima and minima of functions of a single independent variable.

Given the function $f(x)$. Let $h$ be a positive number as small as we please; then the definitions given in §8.4, may be stated as follows: If, for all values of $x$ different from $a$ in the interval $[a - h, a + h]$,

$$f(x)-f(a) =\ {\rm a\ negative\ number},$$ (13.10)

then $f(x)$ is said to be a maximum when $x = a$. If, on the other hand,

$$f(x)-f(a) =\ {\rm a\ positive\ number},$$ (13.11)

then $f(x)$ is said to be a minimum when $x = a$. Consider the following cases:

I

Let $f'(a) \ne 0$. From (13.5), [§13.2], replacing b by x and transposing f(a),

$$f(x)-f(a) = (x-a)f'(x_1),\ \ \ \ a < x_1 < x,$$ (13.12)

Since $f'(a) \ne 0$, and $f'(x)$ is assumed as continuous, $h$ may be chosen so small that $f'(x)$ will have the same sign as $f'(a)$ for all values of $x$ in the interval $[a - h, a + h]$. Therefore $f'(x_1)$ has the same sign as $f'(a)$ (Chap. 3). But $x-a$ changes sign according as $x$ is less or greater than $a$. Therefore, from (13.12), the difference $f(x)-f(a)$ will also change sign, and, by (13.10) and (13.11), $f(a)$ will be neither a maximum nor a minimum. This result agrees with the discussion in §8.4, where it was shown that for all values of $x$ for which $f(x)$ is a maximum or a minimum, the first derivative $f'(x)$ must vanish.

II

Let $f'(a) = 0$, and $f''(a) \ne 0$. From (13.12), replacing $b$ by $x$ and transposing $f(a)$,

$$f(x)-f(a) = \frac{(x - a)^2}{2!} f''(x_2),\ \ \ \ a < x_2 < x.$$ (13.13)

Since $f''(a) \ne 0$, and $f''(x)$ is assumed as continuous, we may choose our interval $[a - h, a + h]$ so small that $f''(x_2)$ will have the same sign as $f''(a)$ (Chap. 3). Also $(x-a)^2$ does not change sign. Therefore the second member of (13.13) will not change sign, and the difference $f(x)-f(a)$ will have the same sign for all values of $x$ in the interval $[a - h, a + h]$, and, moreover, this sign will be the same as the sign of $f''(a)$. It therefore follows from our definitions (13.10) and (13.11) that

$$f(a)\ {\rm is\ a\ maximum\ if}\ f'(a) = 0 \ {\rm and}\ f''(a) =\ {\rm a\ negative\ number};$$ (13.14)

$$f(a)\ {\rm is\ a\ minimum\ if}\ f'(a) = 0 \ {\rm and}\ f''(a) =\ {\rm a\ positive\ number}$$ (13.15)

These conditions are the same as (8.3) and (8.4), [§8.6].

III

Let $f'(a) = f''(a) = 0$, and $f'''(a) \ne 0$. From (13.9), [§13.3], replacing $b$ by $x$ and transposing $f(a)$,

$$f(x)-f(a) = \frac{1}{3!} (x-a)^3 f'''(x_3),\ \ \ \ a < x_3 < x.$$ (13.16)

As before, $f'''(x_3)$ will have the same sign as $f'''(a)$. But $(x-a)^3$ changes its sign from $-$ to $+$ as $x$ increases through $a$. Therefore the difference $f(x)-f(a)$ must change sign, and $f(a)$ is neither a maximum nor a minimum.

IV

Let $f'(a) = f''(a) = \cdots = f^{(n-l)}(a) = 0$, and $f^{(n)}(a) \ne 0$. By continuing the process as illustrated in I, II, and III, it is seen that if the first derivative of $f(x)$ which does not vanish for $x = a$ is of even order ($= n$), then^13.3

$$f(a)\ {\rm is\ a\ maximum\ if}\ f^{(n)}(a) =\ {\rm a\ negative\ number};$$ (13.17)

$$f(a)\ {\rm is\ a\ minimum\ if}\ f^{(n)}(a) =\ {\rm a\ positive\ number}.$$ (13.18)

If the firstderivative of $f(x)$ which does not vanish for $x = a$ is of odd order, then $f(a)$ will be neither a maximum nor a minimum.

Example 13.5.1   Examine $x^3-9x^2 + 24x-7$ for maximum and minimum values.

Solution. $f(x) = x^3-9x^2 + 24x-7$. $f'(x) = 3x^2-18x + 24$. Solving $3x^2 - 18x + 24 = 0$ gives the critical values $x = 2$ and $x = 4$. Thus $f'(2) = 0$, and $f'(4) = 0$. Differentiating again, $f''(x) = 6x-18$. Since $f''(2) =-6$, we know from (13.17) that $f(2) = 13$ is a maximum. Since $f''(4) = + 6$, we know from (13.18) that $f(4) = 9$ is a minimum.

Example 13.5.2   Examine $e^x + 2\cos(x) + e^{-x}$ for maximum and minimum values.

Solution. $f(x) = e^x + 2\cos(x) + e^{-x}$, $f'(x) = e^x-2\sin x-e^{-x} = 0$, for $x = 0$ (and $x = 0$ is the only root of the equation $e^x-2\sin x-e^{-x} = 0$), $f''(x) = e^x - 2\cos(x) + e^{-x} = 0$, for $x = 0$, $f'''(x) = e^x+2\sin x-e^{-x} = 0$, for $x = 0$, $f^{(4)}(x) = e^x + 2\cos(x) + e^{-x} = 4$, for $x = 0$. Hence from (13.18), $f(0) = 4$ is a minimum.