---
title: "Functions and Variations"
book: "High School Mathematics"
subject: math
language: en
chapter: 11
exercises: 11
source: https://one-course.com/books/math/2/en/chapter/11-functions-and-variations
---

# Chapter 11 — Functions and Variations

This chapter organizes what one needs to know about [functions](#def-g11-func-function) before calculus: the small family of *[reference functions](#def-g11-func-reference)* whose [graphs](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) must be known by heart, the precise meaning of *[increasing](#def-g11-func-monotone)* and *[decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations)*, and how variations behave when [functions](#def-g11-func-function) are combined.

## 11.1 Functions and curves

**Definition 11.1 (Function, domain, curve).**

A *function* $f$ assigns to each number $x$ of a set $D \subseteq \R$ (its *domain*) exactly one number $f(x)$. Its *curve* (or [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph)) is the set of points $\bigl(x, f(x)\bigr)$ for $x \in D$.

**Example 11.2.**

The formula $f(x) = \frac{\sqrt{x+2}}{x-1}$ requires $x + 2 \geq 0$ (for the square root) and $x \neq 1$ (nonzero denominator): the [domain](#def-g11-func-function) is $\intco{-2}{1} \cup \intoo{1}{+\infty}$.

## 11.2 The reference functions

The following five [functions](#def-g11-func-function) and their curves must become reflexes.

**Definition 11.3 (Reference functions).**

$$
x \mapsto x^2 \text{ (square)}, \quad
x \mapsto x^3 \text{ (cube)}, \quad
x \mapsto \sqrt{x}, \quad
x \mapsto \frac{1}{x}, \quad
x \mapsto \abs{x}.
$$

Their [domains](#def-g11-func-function) are $\R$, $\R$, $\intco{0}{+\infty}$, $\R \setminus \{0\}$ and $\R$ respectively. The curve of $x \mapsto \frac1x$ is a *hyperbola*.

![The five reference curves: the parabola x2 and the cube x3 (left); √ x, x and the hyperbola 1x (right).](https://one-course.com/images/onecourse/chapters/math-2/g11-func/fig-d927af9605c1.svg)

![The five reference curves: the parabola x2 and the cube x3 (left); √ x, x and the hyperbola 1x (right).](https://one-course.com/images/onecourse/chapters/math-2/g11-func/fig-11cf0cdba8a1.svg)

*The five reference curves: the [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola) $x^2$ and the cube $x^3$ (left); $\sqrt x$, $\abs x$ and the [hyperbola](#def-g11-func-reference) $\frac1x$ (right).*

## 11.3 Variations on an interval

**Definition 11.4 (Increasing, decreasing).**

Let $f$ be defined on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$. The [function](#def-g11-func-function) $f$ is *increasing* on $I$ if it preserves order:

$$
\text{for all } u, v \in I: \quad u < v \implies f(u) \leq f(v),
$$

and *[decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations)* if it reverses order ($u < v \implies f(u) \geq
f(v)$). With strict inequalities on the right, $f$ is *strictly* increasing (resp. [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations)). A [function](#def-g11-func-function) that is increasing or [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $I$ is *monotonic* on $I$.

**Method 11.5 (Proving a variation from the definition).**

Take two arbitrary numbers $u < v$ in the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval), and determine the sign of the difference $f(v) - f(u)$, usually by *[factoring](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand)* it. If the sign is constant on the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval), conclude.

**Proposition 11.6 (Variations of the reference functions).**

1. $x^2$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{-\infty}{0}$ and strictly [increasing](#def-g11-func-monotone) on $\intco{0}{+\infty}$ .
2. $x^3$ is strictly [increasing](#def-g11-func-monotone) on $\R$ .
3. $\sqrt x$ is strictly [increasing](#def-g11-func-monotone) on $\intco{0}{+\infty}$ .
4. $\frac1x$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{-\infty}{0}$ and on $\intoo{0}{+\infty}$ (but *not* on their [union](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) ).
5. $\abs x$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{-\infty}{0}$ and strictly [increasing](#def-g11-func-monotone) on $\intco{0}{+\infty}$ .

**Proof.** Let $u < v$. We factor $f(v) - f(u)$ in each case.

*1.* $v^2 - u^2 = (v-u)(v+u)$. On $\intco{0}{+\infty}$: $v - u > 0$ and $v + u > 0$ (as $v > u \geq 0$), so $v^2 > u^2$. On $\intoc{-\infty}{0}$: $v + u < 0$, so $v^2 < u^2$.

*2.* $v^3 - u^3 = (v - u)(v^2 + uv + u^2)$, and the second factor equals $\left(v + \frac u2\right)^2 + \frac{3u^2}{4} > 0$ unless $u = v = 0$. Hence $v^3 > u^3$ whenever $v > u$.

*3.* For $0 \leq u < v$: $\sqrt v - \sqrt u = \frac{(\sqrt v - \sqrt u)(\sqrt v + \sqrt u)}
{\sqrt v + \sqrt u} = \frac{v - u}{\sqrt v + \sqrt u} > 0$.

*4.* For $0 < u < v$: $\frac1v - \frac1u = \frac{u - v}{uv} < 0$ since $u - v < 0$ and $uv > 0$; same computation for $u < v < 0$. On the [union](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) the claim fails: $-1 < 1$ but $\frac{1}{-1} = -1 < 1 = \frac11$.

*5.* On $\intco{0}{+\infty}$, $\abs x = x$ is strictly [increasing](#def-g11-func-monotone); on $\intoc{-\infty}{0}$, $\abs x = -x$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations). ∎

**Remark 11.7.**

Point 4 is a classic trap: “[decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{-\infty}{0}$ and on $\intoo{0}{+\infty}$” is *not* the same as “[decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\R \setminus \{0\}$”. Variations only make sense on [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval).

**Example 11.8 (Comparing without computing).**

Since $\sqrt x$ is strictly [increasing](#def-g11-func-monotone), $\sqrt{7} < \sqrt{8}$; since $x^2$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{-\infty}{0}$, $(-3)^2 > (-2)^2$. Monotonicity turns comparisons of [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) into comparisons of arguments.

## 11.4 New functions from old

**Proposition 11.9 (Transformations).**

Let $u$ be a [function](#def-g11-func-function) defined on an [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) $I$ and $k \in \R$.

1. $u + k$ has the *same* variations as $u$ on $I$ ; its curve is that of $u$ shifted vertically by $k$ .
2. For $\lambda > 0$ , $\lambda u$ has the same variations as $u$ ; for $\lambda < 0$ , the variations are reversed.
3. If $u \geq 0$ on $I$ , then $\sqrt u$ has the same variations as $u$ .
4. If $u \neq 0$ and $u$ has constant sign on $I$ , then $\frac1u$ has the variations *opposite* to those of $u$ .

**Proof.** Let $u < v$ in $I$ (abusing notation, we compare $x$-values $u < v$; write $s = u(x_1)$, $t = u(x_2)$ for the [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) of $x_1 < x_2$).

*1.* $(u+k)(x_2) - (u+k)(x_1) = t - s$: the difference of [images](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) is unchanged, so its sign is unchanged.

*2.* $\lambda t - \lambda s = \lambda(t - s)$: the sign is kept if $\lambda > 0$, flipped if $\lambda < 0$.

*3.* If $u$ is [increasing](#def-g11-func-monotone) and $x_1 < x_2$, then $0 \leq s \leq t$, and $\sqrt{}$ being [increasing](#def-g11-func-monotone) ([Proposition 11.6](#prop-g11-func-refvariations)) gives $\sqrt s \leq \sqrt t$. Same argument in the [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) case.

*4.* If $u$ is [increasing](#def-g11-func-monotone), $0 < s \leq t$ (or $s \leq t < 0$), then $\frac1s \geq \frac1t$ because $x \mapsto \frac1x$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on each side of $0$: the order is reversed. ∎

**Example 11.10.**

Study $f(x) = \sqrt{x^2 + 1}$ on $\R$. The inner [function](#def-g11-func-function) $u(x) = x^2 + 1$ is positive, [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{-\infty}{0}$ and [increasing](#def-g11-func-monotone) on $\intco{0}{+\infty}$ (a shifted square). By point 3, $f$ has the same variations: [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) then [increasing](#def-g11-func-monotone), with [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f(0) = 1$.

![Transformations in action: shifting x2 up by 1 (red) keeps its variations; taking the square root (orange) keeps them again, as in .](https://one-course.com/images/onecourse/chapters/math-2/g11-func/fig-24fc77b45bdf.svg)

*Transformations in action: shifting $x^2$ up by $1$ (red) keeps its variations; taking the square root (orange) keeps them again, as in [Example 11.10](#ex-g11-func-composite).*

## 11.5 Even and odd functions

**Definition 11.11 (Parity).**

Let $f$ be defined on a [domain](#def-g11-func-function) $D$ symmetric about $0$ (i.e. $-x \in D$ whenever $x \in D$). The [function](#def-g11-func-function) $f$ is *even* if $f(-x) = f(x)$ for all $x \in D$, and *odd* if $f(-x) = -f(x)$ for all $x \in D$.

**Proposition 11.12 (Geometric meaning).**

The curve of an [even function](#def-g11-func-parity) is symmetric about the $y$-axis; the curve of an [odd function](#def-g11-func-parity) is symmetric about the origin.

**Proof.** If $f$ is [even](#def-g11-func-parity) and $(x, y)$ is on the curve (so $y = f(x)$), then $f(-x) = f(x) = y$: the mirror point $(-x, y)$ is also on the curve. If $f$ is [odd](#def-g11-func-parity), $f(-x) = -y$: the point $(-x, -y)$, symmetric of $(x,y)$ about the origin, is on the curve. ∎

**Example 11.13.**

$x^2$ and $\abs x$ are [even](#def-g11-func-parity); $x^3$ and $\frac1x$ are [odd](#def-g11-func-parity); $f(x) = x^2 + x$ is neither: $f(-1) = 0$ while $f(1) = 2$, so $f(-1) \neq \pm f(1)$. Parity halves the work: an [even](#def-g11-func-parity) or [odd function](#def-g11-func-parity) only needs to be studied on $\intco{0}{+\infty}$.

## 11.6 Exercises

**Exercise 11.1 ★.**

Give the [domain](#def-g11-func-function) of each [function](#def-g11-func-function):

$$
f(x) = \sqrt{3 - x}, \qquad
g(x) = \frac{1}{x^2 - 4}, \qquad
h(x) = \frac{\sqrt{x}}{x - 5} .
$$

**Solution of Exercise 11.1.**

$f$: need $3 - x \geq 0$, [domain](#def-g11-func-function) $\intoc{-\infty}{3}$.

$g$: need $x^2 - 4 \neq 0$, i.e. $x \neq \pm 2$: [domain](#def-g11-func-function) $\R \setminus \{-2, 2\}$.

$h$: need $x \geq 0$ and $x \neq 5$: [domain](#def-g11-func-function) $\intco{0}{5} \cup \intoo{5}{+\infty}$.

**Exercise 11.2 ★.**

Using the variations of the [reference functions](#def-g11-func-reference), compare without a calculator:

$$
\sqrt{11} \text{ and } \sqrt{13}; \qquad
(-2.1)^2 \text{ and } (-2.05)^2; \qquad
\frac{1}{0.9} \text{ and } \frac{1}{1.1}; \qquad
(-1.01)^3 \text{ and } (-0.99)^3 .
$$

**Solution of Exercise 11.2.**

$\sqrt{11} < \sqrt{13}$ since $\sqrt{}$ is [increasing](#def-g11-func-monotone) and $11 < 13$.

$(-2.1)^2 > (-2.05)^2$ since $x^2$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{-\infty}{0}$ and $-2.1 < -2.05$.

$\frac{1}{0.9} > \frac{1}{1.1}$ since $\frac1x$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoo{0}{+\infty}$ and $0.9 < 1.1$.

$(-1.01)^3 < (-0.99)^3$ since $x^3$ is [increasing](#def-g11-func-monotone) and $-1.01 < -0.99$.

**Exercise 11.3 ★.**

Determine whether each [function](#def-g11-func-function) is [even](#def-g11-func-parity), [odd](#def-g11-func-parity), or neither:

$$
f(x) = 3x^4 - x^2, \qquad
g(x) = x^3 + x, \qquad
h(x) = x^2 + x^3, \qquad
k(x) = \frac{x}{x^2+1} .
$$

**Solution of Exercise 11.3.**

$f(-x) = 3(-x)^4 - (-x)^2 = 3x^4 - x^2 = f(x)$: [even](#def-g11-func-parity).

$g(-x) = -x^3 - x = -g(x)$: [odd](#def-g11-func-parity).

$h(1) = 2$ and $h(-1) = 1 - 1 = 0$, so $h(-1)$ is neither $h(1)$ nor $-h(1)$: neither [even](#def-g11-func-parity) nor [odd](#def-g11-func-parity).

$k(-x) = \frac{-x}{x^2+1} = -k(x)$: [odd](#def-g11-func-parity).

**Exercise 11.4 ★.**

A [function](#def-g11-func-function) $f$ has the following variation behaviour: [increasing](#def-g11-func-monotone) on $\intcc{-3}{1}$ from $f(-3) = -2$ to $f(1) = 4$, then [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intcc{1}{5}$ down to $f(5) = 0$. How many solutions does the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $f(x) = 1$ have on $\intcc{-3}{5}$? And $f(x) = 5$?

**Solution of Exercise 11.4.**

Read the [variation table](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-table). On $\intcc{-3}{1}$, $f$ climbs from $-2$ to $4$ and crosses the value $1$ exactly once; on $\intcc{1}{5}$ it goes down from $4$ to $0$ and crosses $1$ once more: the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $f(x) = 1$ has $2$ solutions. The [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) of $f$ is $f(1) = 4 < 5$, so $f(x) = 5$ has no solution.

**Exercise 11.5 ★★.**

Using [Method 11.5](#met-g11-func-variation) (sign of $f(v) - f(u)$), prove that $f(x) = x^2 - 4x$ is strictly [increasing](#def-g11-func-monotone) on $\intco{2}{+\infty}$.

**Solution of Exercise 11.5.**

Let $2 \leq u < v$. Then

$$
f(v) - f(u) = v^2 - 4v - u^2 + 4u = (v-u)(v+u) - 4(v-u)
= (v - u)(u + v - 4).
$$

Both factors are positive: $v - u > 0$, and $u + v > 2 + 2 = 4$ since $u \geq 2$ and $v > 2$. Hence $f(v) > f(u)$: $f$ is strictly [increasing](#def-g11-func-monotone) on $\intco{2}{+\infty}$.

**Exercise 11.6 ★★.**

Without computing any derivative, give the variations of

$$
f(x) = \sqrt{5 - x} \ \text{ on } \intoc{-\infty}{5},
\quad
g(x) = \frac{1}{x^2 + 1} \ \text{ on } \intco{0}{+\infty},
\quad
h(x) = 3 - 2\sqrt{x} \ \text{ on } \intco{0}{+\infty}.
$$

**Solution of Exercise 11.6.**

$f = \sqrt{u}$ with $u(x) = 5 - x$ [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) and nonnegative on $\intoc{-\infty}{5}$: $f$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) there ([Proposition 11.9](#prop-g11-func-transforms)).

$g = \frac1u$ with $u(x) = x^2 + 1$ [increasing](#def-g11-func-monotone) and positive on $\intco{0}{+\infty}$: $g$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) there.

$h = 3 - 2\sqrt x$: $\sqrt x$ is [increasing](#def-g11-func-monotone), multiplying by $-2 < 0$ reverses, adding $3$ changes nothing: $h$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intco{0}{+\infty}$.

**Exercise 11.7 ★★.**

Let $f(x) = (x-2)^2 - 3$ on $\R$.

1. Give the variations of $f$ and its [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) .
2. Deduce the variations of $g(x) = \frac{1}{(x-2)^2+1}$ and its [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) . (Note $g = \frac{1}{f + 4}$ .)

**Solution of Exercise 11.7.**

*1.* $f$ is in [canonical form](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#thm-g11-quad-canonical): [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{-\infty}{2}$, [increasing](#def-g11-func-monotone) on $\intco{2}{+\infty}$, [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $f(2) = -3$.

*2.* $g = \frac{1}{u}$ with $u(x) = (x-2)^2 + 1 = f(x) + 4$. The shift by $4$ keeps the variations of $f$, and $u \geq 1 > 0$. Taking the reciprocal reverses them: $g$ is [increasing](#def-g11-func-monotone) on $\intoc{-\infty}{2}$ and [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intco{2}{+\infty}$, with [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $g(2) = \frac{1}{1} = 1$.

**Exercise 11.8 ★★.**

The curve of a [function](#def-g11-func-function) $f$ passes through the points $(0, 1)$, $(2, 3)$ and $(4, 1)$, and $f$ is [increasing](#def-g11-func-monotone) on $\intcc{0}{2}$ and [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intcc{2}{4}$. Sketch a possible curve, then sketch the curves of $f + 2$, of $-f$, and of $2f$.

**Solution of Exercise 11.8.**

Any hill-shaped curve through the three points works. Then: $f + 2$ passes through $(0,3)$, $(2,5)$, $(4,3)$, same variations (vertical shift); $-f$ passes through $(0,-1)$, $(2,-3)$, $(4,-1)$ with *reversed* variations (the peak becomes a valley); $2f$ passes through $(0,2)$, $(2,6)$, $(4,2)$, same variations but amplified vertically.

**Exercise 11.9 ★★.**

Show that the [function](#def-g11-func-function) $f(x) = \frac{2x+1}{x+1}$ can be written $f(x) = 2 - \frac{1}{x+1}$, and deduce its variations on $\intoo{-1}{+\infty}$.

**Solution of Exercise 11.9.**

$2 - \frac{1}{x+1} = \frac{2(x+1) - 1}{x+1} = \frac{2x+1}{x+1} = f(x)$. On $\intoo{-1}{+\infty}$, $u(x) = x + 1$ is [increasing](#def-g11-func-monotone) and positive, so $\frac1u$ is [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations), $-\frac1u$ is [increasing](#def-g11-func-monotone), and adding $2$ changes nothing: $f$ is strictly [increasing](#def-g11-func-monotone) on $\intoo{-1}{+\infty}$.

**Exercise 11.10 ★★.**

True or false (justify or give a counterexample):

1. The sum of two [increasing functions](#def-g11-func-monotone) on $I$ is [increasing](#def-g11-func-monotone) on $I$ .
2. The product of two [increasing functions](#def-g11-func-monotone) on $I$ is [increasing](#def-g11-func-monotone) on $I$ .
3. If $f$ is [increasing](#def-g11-func-monotone) on $I$ , then $f^2$ ( $= f \times f$ ) is [increasing](#def-g11-func-monotone) on $I$ .

**Solution of Exercise 11.10.**

*1. True.* If $u < v$, then $(f+g)(v) - (f+g)(u) = \bigl(f(v)-f(u)\bigr) + \bigl(g(v)-g(u)\bigr)$ is a sum of two nonnegative numbers.

*2. False.* $f(x) = g(x) = x$ are [increasing](#def-g11-func-monotone) on $\R$, but their product $x^2$ is not [increasing](#def-g11-func-monotone) on $\R$ (it decreases on $\intoc{-\infty}{0}$). (The statement becomes true if both [functions](#def-g11-func-function) are also nonnegative.)

*3. False.* Same counterexample: $f(x) = x$ is [increasing](#def-g11-func-monotone) on $\R$ but $f^2(x) = x^2$ is not.

**Exercise 11.11 ★★★.**

Let $f(x) = x + \frac{1}{x}$ for $x > 0$.

1. Show that for $0 < u < v$ , $f(v) - f(u) = (v - u)\,\frac{uv - 1}{uv}$ .
2. Deduce that $f$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{0}{1}$ and strictly [increasing](#def-g11-func-monotone) on $\intco{1}{+\infty}$ .
3. Conclude that $x + \frac1x \geq 2$ for all $x > 0$ , with equality only at $x = 1$ .

**Solution of Exercise 11.11.**

*1.* For $0 < u < v$:

$$
f(v) - f(u) = (v - u) + \frac1v - \frac1u
= (v - u) + \frac{u - v}{uv}
= (v - u)\left(1 - \frac{1}{uv}\right)
= (v - u)\,\frac{uv - 1}{uv}.
$$

*2.* Always $v - u > 0$ and $uv > 0$. If $0 < u < v \leq 1$, then $uv < 1$, so $uv - 1 < 0$ and $f(v) < f(u)$: $f$ is strictly [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{0}{1}$. If $1 \leq u < v$, then $uv > 1$ and $f(v) > f(u)$: $f$ is strictly [increasing](#def-g11-func-monotone) on $\intco{1}{+\infty}$.

*3.* $f$ decreases before $1$ and increases after, so $f(x) \geq f(1) = 2$ for all $x > 0$, with equality exactly at $x = 1$. (This inequality reappears in many guises; compare [Example 10.3](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#ex-g11-quad-canonical) and the tangent-line inequalities of [Chapter 22](https://one-course.com/books/math/2/en/chapter/22-differentiation-and-convexity#ch-g12-deriv).)

## 11.7 Problem: The symmetry factory

**Problem 11.1.**

Weekend problem — every function is an even part plus an odd part: parity rules, the transformation game, and minima won without calculus

Some [functions](#def-g11-func-function) are mirror-symmetric, some are half-turn symmetric, most are neither — yet a small algebraic miracle says that *every* [function](#def-g11-func-function) is, in exactly one way, the sum of a mirror-symmetric piece and a half-turn-symmetric piece. This problem proves the miracle, plays the transformation game on the reference curves ([Proposition 11.9](#prop-g11-func-transforms)), and closes with the [difference-factoring](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand) technique of [Exercise 11.11](#exo-g11-func-11), which wins minimization problems a full chapter before the derivative arrives.

**Part I — Parity fluency.**

1. [Even](#def-g11-func-parity) , [odd](#def-g11-func-parity) , or neither ( [Definition 11.11](#def-g11-func-parity) )? Justify each: $x^4 - 3x^2$ ; $x^3 - x$ ; $x^2 + x$ ; $\abs x$ ; $\frac1x$ .
2. State the geometric meaning of each verdict ( [Proposition 11.12](#prop-g11-func-paritysym) ): which symmetry does the curve of an [even function](#def-g11-func-parity) have? Of an [odd](#def-g11-func-parity) one?
3. Parity multiplies like signs: prove that the product of two [odd functions](#def-g11-func-parity) is [even](#def-g11-func-parity) , and complete the whole multiplication table ( [even](#def-g11-func-parity) $\times$ [even](#def-g11-func-parity) , [even](#def-g11-func-parity) $\times$ [odd](#def-g11-func-parity) , [odd](#def-g11-func-parity) $\times$ [odd](#def-g11-func-parity) ).
4. Quick consequences: if $f$ is [odd](#def-g11-func-parity) and defined at $0$ , what is $f(0)$ ? If $f$ is [even](#def-g11-func-parity) with $f(3) = 7$ , what is $f(-3)$ ? And show that the only [function](#def-g11-func-function) both [even](#def-g11-func-parity) and [odd](#def-g11-func-parity) is the zero [function](#def-g11-func-function) .
5. How much of an [even function](#def-g11-func-parity) ’s curve determines all of it? Give a [function](#def-g11-func-function) that is neither [even](#def-g11-func-parity) nor [odd](#def-g11-func-parity) , and explain what fails.

**Part II — The decomposition theorem.** For a [function](#def-g11-func-function) $f$ defined on a [domain](#def-g11-func-function) symmetric about $0$, set

$$
E(x) = \frac{f(x) + f(-x)}{2},
\qquad
O(x) = \frac{f(x) - f(-x)}{2} .
$$

6. Compute $E$ and $O$ for $f(x) = x^3 + x^2 + 1$ , and check: $E$ [even](#def-g11-func-parity) , $O$ [odd](#def-g11-func-parity) , $E + O = f$ .
7. Prove the general statement: for any $f$ , the [function](#def-g11-func-function) $E$ is [even](#def-g11-func-parity) , $O$ is [odd](#def-g11-func-parity) , and $f = E + O$ — *every [function](#def-g11-func-function) splits into an [even](#def-g11-func-parity) part and an [odd](#def-g11-func-parity) part* .
8. Prove the splitting is unique: if $f = E_1 + O_1 = E_2 + O_2$ with $E_i$ [even](#def-g11-func-parity) and $O_i$ [odd](#def-g11-func-parity) , use question 4 on $E_1 - E_2$ .
9. Decompose $f(x) = \dfrac{1}{1 - x}$ (for $x \neq \pm 1$ ): put $E$ and $O$ over the common denominator $1 - x^2$ and simplify.
10. A look ahead, in one sentence each: the [even](#def-g11-func-parity) and [odd](#def-g11-func-parity) parts of the exponential [function](#def-g11-func-function) will be the twin [functions](#def-g11-func-function) of grade 12 (hyperbolic cosine and sine); and the same instinct — split an object into symmetric pieces and treat each with its own symmetry — powers the university volumes’ analysis of sound and signals. Why is such a split *useful* , in general?

**Part III — The transformation game.**

11. Starting from the curve of $x \mapsto x^2$ , describe precisely the curves of $(x - 3)^2$ , $x^2 + 2$ , $(x - 3)^2 + 2$ , $-x^2$ , $2x^2$ ( [Proposition 11.9](#prop-g11-func-transforms) ).
12. Reduce $f(x) = \abs{x - 2} + \abs{x + 2}$ to a piecewise formula on the three [intervals](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) cut by $-2$ and $2$ , and describe its curve (it has a famous flat bottom).
13. Which transformations respect evenness? Decide, with proof or counterexample: horizontal shift $f(x - 3)$ ; vertical shift $f(x) + 2$ ; vertical stretch $2 f(x)$ (all applied to an [even](#def-g11-func-parity) $f$ ).
14. Solve $\abs{x - 2} + \abs{x + 2} = 6$ using the piecewise formula of question 12.
15. Reverse engineering: a [parabola](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-parabola) has vertex $(2, -3)$ and passes through $(0, 1)$ . Reconstruct its [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) in [canonical form](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#thm-g11-quad-canonical) .

**Part IV — Minima without calculus.**

16. Mirror [Exercise 11.11](#exo-g11-func-11) for $g(x) = x + \dfrac4x$ on $\intoo{0}{+\infty}$ : show $g(v) - g(u) = (v - u)\,\dfrac{uv - 4}{uv}$ , locate the [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) , and give its value.
17. Apply: among all rectangles of area $16$ , which has the least perimeter? (Write $P = 2\left(x +  \frac{16}{x}\right)$ and use question 16’s machine.) An old friend concludes.
18. Determine the variations of $h(x) = \sqrt{x^2 + 1}$ on all of $\R$ , combining a composition argument on $\intco{0}{+\infty}$ with a parity argument for the rest.
19. The closest point: find the point of the curve $y = \sqrt x$ nearest to $(3, 0)$ . (Minimize the *squared* distance — a quadratic in $x$ — and explain why minimizing the square is legitimate.)
20. Finale, the factory tour: reference curves are the raw stock; transformations the machines; parity and the decomposition the quality control; [difference-factoring](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand) the measuring bench. One sentence on each — and name the power tool the next chapter delivers.

**Solution of Problem 11.1.**

**1.** $x^4 - 3x^2$: [even](#def-g11-func-parity) ($(-x)^4 - 3(-x)^2 = x^4 -
3x^2$). $x^3 - x$: [odd](#def-g11-func-parity). $x^2 + x$: neither ($f(-1) = 0 \neq
f(1) = 2$ and $\neq -f(1)$). $\abs x$: [even](#def-g11-func-parity). $\frac1x$: [odd](#def-g11-func-parity).

**2.** [Even](#def-g11-func-parity): the curve is symmetric across the $y$-axis (mirror). [Odd](#def-g11-func-parity): symmetric about the origin (half-turn).

**3.** If $f$ and $g$ are [odd](#def-g11-func-parity): $(fg)(-x) = f(-x)g(-x) = (-f(x))(-g(x)) = (fg)(x)$: [even](#def-g11-func-parity). Table: [even](#def-g11-func-parity) $\times$ [even](#def-g11-func-parity) $=$ [even](#def-g11-func-parity); [even](#def-g11-func-parity) $\times$ [odd](#def-g11-func-parity) $=$ [odd](#def-g11-func-parity); [odd](#def-g11-func-parity) $\times$ [odd](#def-g11-func-parity) $=$ [even](#def-g11-func-parity) — the sign rule with parities.

**4.** [Odd](#def-g11-func-parity) at $0$: $f(0) = -f(0)$, so $f(0) = 0$. [Even](#def-g11-func-parity): $f(-3) = f(3) = 7$. Both [even](#def-g11-func-parity) and [odd](#def-g11-func-parity): $f(x) = f(-x) = -f(x)$ for all $x$, so $2f(x) = 0$: $f$ is the zero [function](#def-g11-func-function).

**5.** The right half ($x \geq 0$) determines an [even function](#def-g11-func-parity) completely: the left half is its mirror. Neither [even](#def-g11-func-parity) nor [odd](#def-g11-func-parity): $x^2 + x$ (question 1) — its two symmetries fail at $x = 1$.

**6.** $E(x) = \frac{(x^3 + x^2 + 1) + (-x^3 + x^2 +
1)}{2} = x^2 + 1$ ([even](#def-g11-func-parity)), $O(x) = \frac{(x^3 + x^2 + 1) -
(-x^3 + x^2 + 1)}{2} = x^3$ ([odd](#def-g11-func-parity)); their sum is $f$.

**7.** $E(-x) = \frac{f(-x) + f(x)}{2} = E(x)$: [even](#def-g11-func-parity). $O(-x) = \frac{f(-x) - f(x)}{2} = -O(x)$: [odd](#def-g11-func-parity). And $E(x) + O(x) = f(x)$ by direct addition: every [function](#def-g11-func-function) splits.

**8.** From $E_1 + O_1 = E_2 + O_2$: $E_1 - E_2 = O_2 - O_1$. The left side is [even](#def-g11-func-parity), the right side [odd](#def-g11-func-parity) — so this [function](#def-g11-func-function) is both, hence zero (question 4): $E_1 = E_2$ and $O_1 = O_2$. One split only.

**9.** $E(x) = \frac12\left(\frac{1}{1 - x} + \frac{1}{1 +
x}\right) = \frac12 \cdot \frac{(1 + x) + (1 - x)}{1 - x^2}
= \frac{1}{1 - x^2}$, and $O(x) = \frac12 \cdot \frac{(1 + x) - (1 - x)}{1 - x^2}
= \frac{x}{1 - x^2}$. Check: $E$ is [even](#def-g11-func-parity), $O$ is [odd](#def-g11-func-parity), and their sum is $\frac{1 + x}{1 - x^2} =
\frac{1 + x}{(1 - x)(1 + x)} = \frac{1}{1 - x} = f(x)$.

**10.** Because each piece obeys a symmetry, half the work (and half the data) suffices for each: solve on one side, mirror to the other — and many operations (integration, series, signal analysis) treat symmetric pieces far more simply than raw [functions](#def-g11-func-function).

**11.** $(x-3)^2$: shifted $3$ right. $x^2 + 2$: shifted $2$ up. $(x-3)^2 + 2$: both. $-x^2$: flipped upside down. $2x^2$: stretched vertically by $2$ (narrower bowl).

**12.** For $x \geq 2$: $(x - 2) + (x + 2) = 2x$. For $-2 \leq x \leq 2$: $(2 - x) + (x + 2) = 4$. For $x \leq -2$: $-2x$. The curve descends with [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $-2$, runs flat at height $4$ between $-2$ and $2$, then climbs with [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $2$: a valley with a flat floor.

**13.** Vertical shift: $f(x) + 2$ stays [even](#def-g11-func-parity) ($f(-x) + 2 = f(x) + 2$). Vertical stretch: $2f(x)$ stays [even](#def-g11-func-parity). Horizontal shift: $(x - 3)^2$ is not [even](#def-g11-func-parity) ($f(-1) = 16 \neq 4 = f(1)$): sliding sideways destroys the mirror.

**14.** On the flat floor the value is $4 \neq 6$. For $x \geq 2$: $2x = 6$, $x = 3$. For $x \leq -2$: $-2x = 6$, $x = -3$. Solutions: $\pm 3$.

**15.** $f(x) = a(x - 2)^2 - 3$ with $f(0) = 4a - 3 = 1$: $a = 1$, so $f(x) = (x - 2)^2 - 3$.

**16.** $g(v) - g(u) = (v - u) + 4\,\frac{u - v}{uv}
= (v - u)\,\frac{uv - 4}{uv}$: negative for $u < v$ with $uv < 4$, positive with $uv > 4$: [decreasing](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-variations) on $\intoc{0}{2}$, [increasing](#def-g11-func-monotone) on $\intco{2}{+\infty}$; [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) at $x = 2$, of value $g(2) = 4$.

**17.** Sides $x$ and $\frac{16}{x}$: $P(x) = 2\left(x + \frac{16}{x}\right)$, minimized (question 16’s machine with $4 \to 16$, turning point $\sqrt{16} = 4$) at $x = 4$: the $4 \times 4$ *square*, perimeter $16$ — Dido’s answer, now proved for every real side length.

**18.** On $\intco{0}{+\infty}$: $x \mapsto x^2 + 1$ increases, and $\sqrt{\phantom{x}}$ increases, so the composite increases. And $h(-x) = \sqrt{x^2 + 1} = h(x)$: $h$ is [even](#def-g11-func-parity), so by mirror symmetry it decreases on $\intoc{-\infty}{0}$. [Minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $h(0) = 1$.

**19.** Squared distance to $(3, 0)$ of the point $(x, \sqrt x)$: $d^2(x) = (x - 3)^2 + x = x^2 - 5x + 9$, minimal at $x = \frac52$: closest point $\left(\frac52, \sqrt{\frac52}\right)$, distance $\sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}$. Minimizing $d^2$ is legitimate because squaring is strictly [increasing](#def-g11-func-monotone) on positive numbers: $d$ and $d^2$ bottom out at the same spot.

**20.** Raw stock: the reference curves, known by heart. Machines: shifts, flips and stretches assembling new [graphs](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) from old. Quality control: parity and the even–odd split, which halve every study. Measuring bench: [factoring](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand) $f(v) - f(u)$, which reads variations and minima directly. The next chapter’s power tool: the *derivative*, which reads them at a glance.