Week 1 · Day 5 Lesson 1.12–1.13 About 13 minutes Free

Monotonicity and the same-function test

By the end: Find where a function is increasing or decreasing, use monotonicity to solve inequalities and find parameters, and decide whether two functions are really the same.

Understand it

A function is increasing on an interval if moving right makes the output grow: for $x_1<x_2$ in the interval, $f(x_1)<f(x_2)$ . It is decreasing if the output shrinks: $x_1<x_2$ gives $f(x_1)>f(x_2)$ . Decreasing does not mean negative values; it means the graph goes downhill from left to right.

Always name the interval. Monotonicity belongs to an interval, not to a function as a whole. For $f(x)=\dfrac1x$ , the function is decreasing on $(-\infty,0)$ and decreasing on $(0,+\infty)$ . It is not decreasing on the union $(-\infty,0)\cup(0,+\infty)$ , because $f(-1)=-1<f(1)=1$ even though $-1<1$ . List separate intervals separately, with “and” or a comma, never with $\cup$ .

Common functions. A line $y=kx+m$ is increasing for $k>0$ and decreasing for $k<0$ . A parabola $y=ax^2+bx+c$ with $a>0$ is decreasing on $\left(-\infty,-\dfrac b{2a}\right]$ and increasing on $\left[-\dfrac b{2a},+\infty\right)$ ; for $a<0$ it is the other way round.

Using monotonicity. If $f$ is increasing, then $f(m)<f(n)\iff m<n$ : drop $f$ and keep the direction. If $f$ is decreasing, drop $f$ and reverse the direction. Remember that $m$ and $n$ must lie in the domain, so write those conditions as well.

Parameter questions. “ $f$ is increasing on $[2,+\infty)$ ” means that interval must fit inside the function’s increasing part. For a parabola, compare the vertex $x$ with $2$ .

The same function. Two functions are the same only if they have the same domain and the same rule. The name of the variable does not matter ( $f(x)=x+1$ and $g(t)=t+1$ are the same). Check the domain first, then the rule. A formula that simplifies is not enough: $\dfrac{x^2}{x}$ simplifies to $x$ , but it is not defined at $0$ , so it is a different function from $x$ .

Essential formulas

x_1<x_2\ \Rightarrow\ f(x_1)<f(x_2)\ \ (\text{increasing}),\qquad f(x_1)>f(x_2)\ \ (\text{decreasing})

These must hold for all

x_1<x_2

in the interval you name.

y=ax^2+bx+c,\ a>0:\ \text{decreasing on }\left(-\infty,-\tfrac b{2a}\right],\ \text{increasing on }\left[-\tfrac b{2a},+\infty\right)

The turning point is the vertex at

x=-\dfrac b{2a}

. Both intervals include the vertex.

f\ \text{increasing: }f(m)<f(n)\iff m<n\qquad f\ \text{decreasing: }f(m)<f(n)\iff m>n

Add the conditions that

m

and

n

are in the domain.

f=g\iff\text{same domain and same rule}

The letter used for the variable is irrelevant. A different domain makes it a different function.

See it

Decreasing on (−∞, 2], increasing on [2, +∞). The turning point is the vertex (2, −3).

Worked examples

Worked example 1

Find the intervals where $f(x)=x^2-4x+1$ is increasing and decreasing.

The parabola opens upward ( $a=1>0$ ). Its vertex is at $x=-\dfrac{b}{2a}=-\dfrac{-4}{2}=2$ .
Left of the vertex the graph falls, so $f$ is decreasing on $(-\infty,2]$ .
Right of the vertex it rises, so $f$ is increasing on $[2,+\infty)$ .

Answer: Decreasing on $(-\infty,2]$ , increasing on $[2,+\infty)$ .

Worked example 2

$f$ is decreasing on $[-2,2]$ and $f(m-1)>f(2m-1)$ . Find the range of $m$ .

Both inputs must be in the domain $[-2,2]$ : $-2\le m-1\le2$ gives $-1\le m\le3$ , and $-2\le2m-1\le2$ gives $-\tfrac12\le m\le\tfrac32$ .
Since $f$ is decreasing, drop $f$ and reverse the direction: $f(m-1)>f(2m-1)\iff m-1<2m-1$ , so $m>0$ .
All three conditions together: $m>0$ and $-\tfrac12\le m\le\tfrac32$ (and $-1\le m\le3$ is automatic).

Answer: $0<m\le\tfrac32$

Worked example 3

Which pairs are the same function? (a) $f(x)=\dfrac{x^2}{x}$ and $g(x)=x$ ; (b) $f(x)=|x|$ and $g(x)=\sqrt{x^2}$ ; (c) $f(x)=x+1$ and $g(t)=t+1$ ; (d) $f(x)=\sqrt{x+1}\cdot\sqrt{x-1}$ and $g(x)=\sqrt{x^2-1}$ .

(a) The domain of $f$ excludes $0$ , but the domain of $g$ is $\mathbb{R}$ . Different.
(b) Both have domain $\mathbb{R}$ , and $\sqrt{x^2}=|x|$ for every $x$ . Same.
(c) The letter does not matter: same domain $\mathbb{R}$ and same rule. Same.
(d) $f$ needs $x+1\ge0$ and $x-1\ge0$ , so its domain is $x\ge1$ . $g$ needs $x^2-1\ge0$ , so its domain is $x\le-1$ or $x\ge1$ . Different domains.

Answer: (b) and (c) are the same function.

Try it yourself

Work it on paper first. Open a hint only when you are stuck, then compare with the solution.

The function $f(x)=x^2+2(a-1)x+2$ is decreasing on $(-\infty,4]$ . Find the range of $a$ .

Hint 1 of 2

Find the vertex $x$ in terms of $a$ . The parabola opens upward.

Hint 2 of 2

The interval $(-\infty,4]$ must fit inside the decreasing part $(-\infty,\text{vertex}]$ .

Show the step-by-step solution

The vertex is at $x=-\dfrac{2(a-1)}{2}=1-a$ .
The parabola opens upward, so $f$ is decreasing on $(-\infty,1-a]$ .
For $f$ to be decreasing on $(-\infty,4]$ , we need $4\le1-a$ , so $a\le-3$ .

Answer: $a\le-3$

Common mistakes

Joining separate intervals with $\cup$ . $\dfrac1x$ is decreasing on $(-\infty,0)$ and on $(0,+\infty)$ , not on their union.
Forgetting to reverse the inequality when $f$ is decreasing.
Ignoring the domain when solving $f(m)<f(n)$ . The inputs $m$ and $n$ must be allowed.
Calling two functions the same because their simplified formulas match. Compare the domains first.
Believing “decreasing” means the values are negative. It describes the direction of the graph.

Exam tips

For a parabola, find the vertex $x=-\dfrac b{2a}$ first; it splits the increasing and decreasing parts.
For a parameter question, put the given interval on a number line next to the function’s own interval and compare the endpoints.
For “same function”, write the domain of each formula first. If they differ, you are done.
When you drop $f$ from an inequality, write the domain conditions on the same line so you do not forget them.

In short

Monotonicity is always on an interval; list separate intervals separately.
Increasing: keep the direction when dropping $f$ . Decreasing: reverse it.
A parabola changes direction at its vertex $x=-\dfrac b{2a}$ .
Same function = same domain and same rule.

Knowledge check

4 questions on this lesson, marked as you go. Take them in ACE CSCA with a free account: passing them completes the lesson, and your progress is saved on every device.

All of Week 1 is free: the eight lessons, the daily questions and the Week 1 test.

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