Chapter 6 · Application of Derivatives

Example 10: Where $x^2-4x+6$ increases and decreases

ExamplesIncreasing and decreasing functions · Section 6.3

The question

Find the intervals in which the function $f$ given by $f(x) = x^2 - 4x + 6$ is

(a) increasing (b) decreasing

Exam answer

f'(x) = 2x - 4 = 2(x - 2)

$f'(x) = 0 \Rightarrow x = 2$ ⁠. This point divides $\mathbf{R}$ into $(-\infty, 2)$ and $(2, \infty)$ ⁠.

In $(-\infty, 2)$ ⁠, $f'(x) < 0$ ⁠.

In $(2, \infty)$ ⁠, $f'(x) > 0$ ⁠.

(a) $f$ is increasing in $(2, \infty)$ ⁠.

(b) $f$ is decreasing in $(-\infty, 2)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The direction of the graph is the sign of the derivative. $f'(x) = 2x - 4 = 2(x - 2)$ is zero only at $x = 2$ ⁠, negative to its left and positive to its right. So the point 2 splits the real line into two intervals: on one the function falls, on the other it rises. The graph is a parabola with its lowest point at $x = 2$ ⁠.

Given, to find, and the plan

Given: $f(x) = x^2 - 4x + 6$ for every real $x$ (a polynomial, so continuous and differentiable everywhere).
To find: (a) the intervals where $f$ is increasing; (b) the intervals where $f$ is decreasing.
Plan: Differentiate; solve $f'(x) = 0$ ⁠; make a sign chart of $f'$ on the two intervals; conclude with Theorem 1; check by completing the square and by values.

Basics you need

1Definition 1 and Theorem 1 (Section 6.3)

$f$ is increasing on I if $x_1 < x_2$ in I $\Rightarrow f(x_1) \le f(x_2)$ (strictly: $<$ ⁠), decreasing if $\Rightarrow f(x_1) \ge f(x_2)$ (strictly: $>$ ⁠). Theorem 1: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠, then $f$ is increasing in $[a, b]$ if $f' \ge 0$ on $(a, b)$ ⁠, and decreasing if $f' \le 0$ there. Lesson: increasing and decreasing functions

2Sign of a linear factor

$x - 2$ is negative when $x < 2$ ⁠, zero at $x = 2$ ⁠, positive when $x > 2$ ⁠; multiplying by the positive number 2 does not change the sign. Basics: sign of a product

3Interval notation

$(-\infty, 2)$ means all $x < 2$ ⁠; $(2, \infty)$ means all $x > 2$ ⁠; $(-\infty, 2]$ includes 2. Basics: interval notation

Solution

Differentiate and find where $f'(x) = 0$
$f'(x) = 2x-4 = 2(x - 2) .$
So $f'(x) = 0$ exactly when $x = 2$ ⁠.

The sign of $f'$ decides where $f$ rises or falls, and the factor form $2(x - 2)$ shows that the sign can switch only at the zero $x = 2$ ⁠. This point divides the real line into the two disjoint intervals $(-\infty, 2)$ and $(2, \infty)$ ⁠; Step 2 shows that $f'$ keeps one sign on each.

Power rule: $\frac{d}{dx}(x^2) = 2x$ ⁠, $\frac{d}{dx}(4x) = 4$ ⁠, $\frac{d}{dx}(6) = 0$ (Section 5.3).

Sign chart of $f'$

Interval	$x - 2$	$f'(x) = 2(x-2)$	Test value
$(-\infty,2)$	$-$	$-$	$f'(0) = -4$
$(2,\infty)$	$+$	$+$	$f'(3) = 2$

Sign chart (not to scale). The dashed line is at

x = 2

⁠, where

f'(x) = 0

⁠; left of it

f'

is negative, right of it positive.

For

x < 2

the factor

x - 2

is negative, so

f'(x) = 2(x - 2) < 0

at every point of

(-\infty, 2)

⁠, not just at the test value. For

x > 2

it is positive. The test values confirm each row with an actual number.

Basics: sign of a product

Conclude with Theorem 1

On $(-\infty, 2)$ ⁠. For any $x_1 < x_2$ in this interval, $f$ is continuous on $[x_1, x_2]$ ⁠, differentiable on $(x_1, x_2)$ ⁠, and $f' < 0$ there. By Theorem 1(b), $f(x_1) \ge f(x_2)$ ⁠; the Mean Value Theorem argument of the book's proof even gives $f(x_1) > f(x_2)$ ⁠. So $f$ is decreasing on $(-\infty, 2)$ .

On $(2, \infty)$ ⁠. In the same way $f' > 0$ there, so by Theorem 1(a) $f$ is increasing on $(2, \infty)$ , in fact strictly.

This is the book's answer, with the open intervals on which $f'$ is strictly negative or positive. Theorem 1 is applied between any two points of each interval, exactly as in Example 9.

Theorem 1, Section 6.3. Lesson: increasing and decreasing functions
The point $x = 2$ can be included

Completing the square gives the whole picture:
$f(x) = x^2 - 4x + 4 + 2 = (x-2)^2 + 2 .$
For any real $a < b$ ⁠,
$f(b) - f(a) = (b - a)(a + b - 4) .$
If $a < b \le 2$ then $a + b < 4$ ⁠, so $f(b) < f(a)$ ⁠; if $2 \le a < b$ then $a + b > 4$ ⁠, so $f(b) > f(a)$ ⁠. Hence $f$ is strictly decreasing on $(-\infty, 2]$ and strictly increasing on $[2, \infty)$ ⁠.

The graph of $y = x^2 - 4x + 6$ (not to scale: the $y$ unit is drawn shorter than the $x$ unit). It falls until its lowest point $(2, 2)$ and rises after it.

The graph has a horizontal tangent at $x = 2$ but no break, so 2 can be added to either interval: $f$ is continuous there. The book's open intervals are correct; the closed ones say slightly more. What we may not do is say $f$ is increasing on all of $\mathbf{R}$ ⁠: $0 < 2$ but $f(0) = 6 > 2 = f(2)$ ⁠.

$b^2 - a^2 = (b - a)(b + a)$ ⁠, so $f(b) - f(a)$ $= (b^2 - a^2) - 4(b - a)$ $= (b - a)(a + b - 4)$ ⁠.

Result

$f'(x) = 2x - 4$ ⁠, which is zero only at $x = 2$ ⁠.

(a) $f$ is increasing on $(2, \infty)$ . (b) $f$ is decreasing on $(-\infty, 2)$ .

Both are strict, and the point 2 may be included: strictly decreasing on $(-\infty, 2]$ ⁠, strictly increasing on $[2, \infty)$ ⁠. This agrees with the book's solution.

Check it with numbers

$x$	$0$	$1$	$2$	$3$	$4$
$f(x)$	6	3	2	3	6

Down from 6 to 2 as $x$ goes from 0 to 2, then up from 2 to 6 as $x$ goes from 2 to 4: exactly the sign chart. The symmetry about $x = 2$ comes from $f(x) = (x - 2)^2 + 2$ ⁠.

Common mistakes

Stopping at $f'(x) = 0$ ⁠. $x = 2$ is only the boundary; the answer is the sign of $f'$ on each side.
Swapping the answers. Negative $f'$ means decreasing. Check with values: $f(0) = 6 > f(1) = 3$ ⁠, so $f$ falls left of 2.
Calling $f$ increasing on $\mathbf{R}$ because the parabola "opens upward". Opening upward means it falls and then rises.

What to take away

Recipe: differentiate, find where $f' = 0$ ⁠, sign chart, then $+$ means increasing and $-$ means decreasing.
For an upward parabola $ax^2 + bx + c$ ( $a > 0$ ⁠), the turning point is at $x = -\frac{b}{2a}$ ⁠: decreasing before it, increasing after it.
A boundary point where $f$ is continuous may be added to the intervals.

Every fact used, and where it comes from

Fact	Used in	Source
Power rule; derivative of a constant is 0	Step 1	Table 5.3 and the algebra of derivatives, Section 5.3
Sign of the linear factor $x - 2$	Step 2	Basics: sign of a product
Interval notation	Steps 2, 3, 4	Basics: interval notation
Sign of $f'$ decides increasing or decreasing	Step 3	Theorem 1 (a), (b), Section 6.3 (Lesson)
Increasing, decreasing, strict versions	Steps 3, 4	Definition 1, Section 6.3
Mean Value Theorem (for strictness)	Step 3	Used without proof in the book's proof of Theorem 1, which cites it as "Theorem 8 in Chapter 5"; this reprint's Chapter 5 does not contain it (a stale reference). Step 4 proves strictness without it.
Completing the square; $b^2 - a^2 = (b-a)(b+a)$	Step 4	Basics: factorising and the discriminant
Table of values	Check	Arithmetic, recomputed in the site's verification script

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Example 10: Where x2−4x+6x^2-4x+6x2−4x+6 increases and decreases

Basics you need

Solution

Differentiate and find where f′(x)=0f'(x) = 0f′(x)=0

Sign chart of f′f'f′

Conclude with Theorem 1

The point x=2x = 2x=2 can be included

Every fact used, and where it comes from

Example 10: Where $x^2-4x+6$ increases and decreases

Differentiate and find where $f'(x) = 0$

Sign chart of $f'$

The point $x = 2$ can be included