Exercise 6.3 · Q4: Why $e^x$⁠, $\log x$ and $x^3+x^2+x+1$ have no maxima or minima

Basics you need

Part (i): $f(x)=e^x$

1. A maximum or minimum value would be a local one

   This step serves all three parts. Suppose a function $F$ on an open interval I had a maximum value at some $c\in$ I: $F(c)\ge F(x)$ for every $x\in$ I. Because I is open, there is an $h\gt0$ with $(c-h,c+h)$ inside I, and the inequality holds in particular for those $x$⁠. That is exactly Definition 4(a): $c$ is a point of local maxima. The same holds for minimum values.

   So once we show there are no points of local maxima or minima, there can be no maximum or minimum values either. The domains here, $\R$ and $(0,\infty)$⁠, are open intervals, so this applies.
   Definitions 3 and 4, Section 6.4. Lesson: maxima and minima

2. $f'(x)=e^x$ is never zero

   $f$ is differentiable at every real $x$⁠, with $f'(x)=e^x$⁠. An exponential with a positive base is always positive, so $f'(x)=e^x\gt0\ne0$ for every $x$⁠.

   By Theorem 2, a point of local maxima or minima would need $f'(c)=0$ or $f$ not differentiable at $c$⁠. Neither ever happens, so $e^x$ has no local maxima or minima, and by Part (i), Step 1, no maximum or minimum value.

   There are no critical points at all. Theorem 2 is a necessary condition: no critical point means nowhere for an extreme to be.
   Theorem 2, Section 6.4. Lesson: tests

3. Direct confirmation

   For any $c$⁠: $f(c+1)=e\cdot e^c\gt e^c=f(c)$ and $f(c-1)=e^c/e\lt e^c$⁠, because $e\gt1$⁠. Every candidate is beaten on both sides. The values $e^x$ get as close to $0$ as we like (for example $e^{-10}\approx0.0000454$⁠) but never reach it, just as in Example 16.

   This shows the conclusion without Theorem 2, which is a good check that we read the theorem in the right direction.

Part (ii): $g(x)=\log x$

1. $g'(x)=\frac1x$ is never zero on $(0,\infty)$

   The domain of $\log x$ is $(0,\infty)$⁠, an open interval. On it $g$ is differentiable with $g'(x)=\frac1x\gt0$⁠. A fraction with numerator $1$ is never zero. So there is no critical point, and by Theorem 2 $\log x$ has no local maxima or minima; by Part (i), Step 1, no maximum or minimum value.

   The point $x=0$ might look special, but it is not in the domain, so it cannot be a point of maxima or minima.
   $\frac{d}{dx}\log x=\frac1x$⁠, Chapter 5.

2. Direct confirmation

   For any $c\gt0$⁠: $\log(2c)=\log2+\log c\gt\log c$ and $\log\frac{c}{2}=\log c-\log2\lt\log c$⁠, since $\log2\gt0$⁠. Both $2c$ and $\frac c2$ are positive, so they are in the domain.

   Doubling or halving keeps us inside $(0,\infty)$⁠, unlike adding or subtracting $1$⁠, which could leave the domain when $c$ is small.
   $\log(ab)=\log a+\log b$⁠.

Part (iii): $h(x)=x^3+x^2+x+1$

1. $h'(x)=3x^2+2x+1$ is always positive

   $h$ is a polynomial, differentiable everywhere, with $h'(x)=3x^2+2x+1$⁠. Its discriminant is $2^2-4\cdot3\cdot1=-8\lt0$ and its leading coefficient $3$ is positive, so $h'(x)\gt0$ for every $x$⁠. Without the discriminant:

   $$3x^2+2x+1=2x^2+(x+1)^2.$$

   Both terms are $\ge0$⁠, and they cannot both be zero (that would need $x=0$ and $x=-1$ at once), so the sum is $\gt0$⁠.

   Expanding: $2x^2+(x^2+2x+1)=3x^2+2x+1$⁠. ✓ Completing the square gives the exact least value: $3\left(x+\frac13\right)^2+\frac23\ge\frac23$⁠.
   Negative discriminant ⇒ no real roots. Basics: the discriminant

2. Conclude with Theorem 2

   $h'(c)\ne0$ for every $c$⁠, and $h$ is differentiable everywhere, so there is no critical point. By Theorem 2, $h$ has no local maxima or minima, and by Part (i), Step 1, no maximum or minimum value on $\R$⁠.

   Since $h'\gt0$ everywhere, $h$ is in fact strictly increasing on $\R$ (Theorem 1 and the Mean Value step of its proof, applied on $[x_1,x_2]$ for any $x_1\lt x_2$⁠), so every point is beaten by any point to its right and undercuts any point to its left.
   Theorem 2, Section 6.4; Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

Every fact used, and where it comes from