MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: monotonic functions and the First Derivative Test (MATH 203)

Q: How do I find where a function is increasing or decreasing in MATH 203?

Differentiate, then rewrite the derivative as one fraction with a completely factored numerator and denominator. List the numbers where it is zero or undefined, and make a sign chart with one row per factor. Where the derivative is positive the function is increasing, where it is negative it is decreasing, interval by interval, never across a point outside the domain.

Q: What is the First Derivative Test for local extrema?

At a critical point where the function is continuous, look at the sign of the derivative just before and just after. From negative to positive gives a local minimum, from positive to negative a local maximum, and no change of sign means no local extremum. The test needs no second derivative, so it works at cusps and corners too.

Q: Why does my sign chart give the wrong intervals when my derivative looks right?

Usually because the derivative was never fully factored, or a step lost information: a minus sign of the quotient rule applied to one term only, a cosine or a power of x divided away, or an inequality multiplied by a denominator that can be negative. Test the simplified derivative at two easy points against the raw one before building the chart.

Q: Can an endpoint of an interval be a local maximum or minimum?

Yes, in Thomas's definition used in MATH 203. Compare the value at the endpoint with the values of the function at nearby points of the domain. If the function decreases right after the left endpoint, that endpoint is a local maximum; if it increases right after, a local minimum. On a closed interval both ends belong on the list of local extrema.

Q: Is a function still increasing if its derivative is zero at one point?

Yes. If the derivative is positive on both sides of an isolated zero, the function is increasing on each side and the two intervals join at that point, so it is increasing on the whole interval. The cube function is the standard example: its derivative vanishes at the origin, and it still increases everywhere, with no extremum.

This sheet is not a summary of section 4.3 of Thomas' Calculus: you have the course notes. It answers one question for MATH 203 at Concordia University: where do students lose marks when they find where a function increases and decreases and classify its critical points, and which precise gesture avoids each loss.

The test fits in three lines and every student can recite it. The marks go in the algebra that has to happen first: $f'$ must become ONE fraction whose numerator and denominator are factored, because the sign of a product is read factor by factor and the sign of a sum is not read at all. Numbers below are exact; a scientific calculator is allowed on the exam, but it cannot factor for you.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

The First Derivative Test is a question of SIGN, and a sign is only readable on $f'$ written as ONE fraction, FULLY factored: the marks are lost in the algebra that leads there, a minus sign not distributed, a $\cos x$ divided away, a lowest power not factored out, a point outside the domain kept or forgotten.

This chapter is part of MATH 203, Calculus I (Concordia)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Increasing and decreasing functions
Monotonic function
Critical point
First Derivative Test for Local Extrema
Local extremum at an endpoint
Sign chart
Corollary of the Mean Value Theorem
Quotient rule
Rational exponents
One-to-one function

The essentials

One fraction, fully factored, then the chart

• The sign of a SUM is unreadable; the sign of a PRODUCT or QUOTIENT is read factor by factor. So $f'$ is always brought to the form $\frac{\text{factored numerator}}{\text{factored denominator}}$ before any sign is written.
• A factor of odd power, $(x - a)$ , $(x - a)^3$ , $x^{1/3}$ , changes sign at $a$ . A factor of even power, $(x - a)^2$ , $x^{2/3}$ , does not.
• A factor that is never zero on the domain, $e^x$ , $x^2 + 1$ , $x^2 + x + 1$ (discriminant $-3$ ), $\sqrt{6 - x}$ inside its domain, is dropped from the SIGN, never from the derivative.
• The chart has one column per zero of $f'$ AND one per number where $f'$ is undefined, including the numbers outside the domain of $f$ , marked with a double bar: those get no verdict.

f'(x) = \frac{(x + 1)(x - 2)}{(x - 1)^2}

: the squared factor stays positive, and at

1

, outside the domain,

f'

keeps its sign and nothing is concluded.

Marker's habit: a chart whose rows are the factors, as in the figure, gets full marks for the signs even if a value is slipped later; a single row of signs with no justification gets nothing when one sign is wrong.

What the test needs, and what it concludes

• The corollary: $f$ continuous on $[a, b]$ and $f' > 0$ on $(a, b)$ give $f$ increasing on $[a, b]$ . Continuity at the ENDPOINTS is required; differentiability there is not.
• Critical point: $c$ IN THE DOMAIN with $f'(c) = 0$ or $f'(c)$ undefined. A number outside the domain is never a critical point.
• First Derivative Test: $f$ continuous at $c$ ; $-$ to $+$ gives a local minimum, $+$ to $-$ a local maximum, no sign change no extremum.
• On a closed interval, the endpoints carry local extrema too (Thomas's definition): decreasing right after $a$ makes $f(a)$ a local maximum.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

How each kind of factor behaves on the sign chart

Read a line as: when $f'$ contains the factor of the first column, its sign across the zero is in the second, and what the chart concludes about $f$ is in the third. The red lines are not conclusions: they are steps that destroy information.

Factor of $f'$	Across its zero	For $f$
$x - a$	changes	extremum at $a$
Example: $f'(x) = 3(x - 4)(x + 2)$ : $-$ to $+$ at $4$ , local minimum $f(4)$ for $f(x) = x^3 - 3x^2 - 24x + 4$ , equal to $-76$ .
$(x - a)^2$	keeps	none at $a$
Example: $h'(x) = 5x^2(x - 1)(x - 3)$ : $h'(-1) = 40 > 0$ and $h'(0.5) > 0$ , no extremum at $0$ .
$x^{1/3}$ below	changes	extremum, cusp
Example: $f'(x) = \frac{8(x - 2)(x + 2)}{3x^{1/3}}$ : $+$ to $-$ at $0$ , local maximum $f(0) = 0$ at a cusp.
$e^x$ , $x^2 + x + 1$	never zero	drop from sign
Example: $g'(x) = 12x(x^2 + x + 1)$ : discriminant $1 - 4 = -3$ , the sign is that of $12x$ , minimum $g(0) = -5$ .
$(x - a)^3$ below	changes	a not in domain no verdict at a
Example: $g'(x) = -\frac{x + 2}{(x - 2)^3}$ : $+$ to $-$ at $2$ , and $g(x) = \frac{x}{(x - 2)^2}$ has no maximum. Same form, other result: $\frac{8(x - 2)(x + 2)}{3x^{1/3}}$ changes from $+$ to $-$ at $0$ too, and there $0$ IS in the domain: a maximum. What to do: Double bar at $a$ , intervals stated on each side separately, no extremum.
$\cos x$ divided out	zeros lost	missing points illegal step
Example: $2\cos x(1 - 2\sin x) = 0$ divided by $\cos x$ keeps $\frac{\pi}{6}$ , $\frac{5\pi}{6}$ and loses $\frac{\pi}{2}$ , $\frac{3\pi}{2}$ , where $f = -3$ . What to do: Factor, then set EACH factor equal to $0$ on the interval.

The first four lines are the whole chapter; the last two are the two ways a correct derivative still produces a wrong answer.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Applying the minus sign of the quotient rule to one term only

the whole question: every critical point is wrong

What not to write

“ $f'(x) = \frac{(2x + 5)(x - 4) - x^2 + 5x}{(x - 4)^2} = \frac{x^2 + 2x - 20}{(x - 4)^2}$ .”

What to write

“ $f'(x) = \frac{(2x + 5)(x - 4) - (x^2 + 5x)}{(x - 4)^2} = \frac{x^2 - 8x - 20}{(x - 4)^2} = \frac{(x - 10)(x + 2)}{(x - 4)^2}$ .”

Why: The minus sign multiplies the whole product $uv'$ . Brackets around $uv'$ BEFORE expanding cost nothing; forgetting them turns $-5x$ into $+5x$ and the chart into fiction.

2. Dividing by cos x, or by x, to solve f'(x) = 0

2 to 3 marks, and often the lowest point of the graph

What not to write

“ $-4\sin x\cos x + 2\cos x = 0$ , so dividing by $2\cos x$ : $\sin x = \frac{1}{2}$ , the critical points are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ .”

What to write

“ $2\cos x(1 - 2\sin x) = 0$ , so $\cos x = 0$ or $\sin x = \frac{1}{2}$ : the critical points in $(0, 2\pi)$ are $\frac{\pi}{6}$ , $\frac{\pi}{2}$ , $\frac{5\pi}{6}$ and $\frac{3\pi}{2}$ .”

Why: Dividing by an expression is only allowed where it is not zero, and its zeros are exactly solutions of the equation. Here $f\left(\frac{3\pi}{2}\right) = -3$ is the lowest value, lost in one line.

3. Multiplying through by a fractional power and losing the cusp

2 marks: the two minima of the graph

What not to write

“ $f(x) = (x^2 - 1)^{2/3}$ , $f'(x) = \frac{4x}{3}(x^2 - 1)^{-1/3} = 0$ ; multiplying by $(x^2 - 1)^{1/3}$ : $x = 0$ , the only critical point.”

What to write

“ $f'(x) = \frac{4x}{3(x^2 - 1)^{1/3}}$ : critical points $0$ ( $f' = 0$ ) and $\pm 1$ ( $f'$ undefined, $f(\pm 1) = 0$ ). Local minima $0$ at $\pm 1$ , local maximum $f(0) = 1$ .”

(x^2 - 1)^{2/3}

: the equation

f'(x) = 0

finds only the maximum at

0

; the two minima are cusps at

x = \pm 1

, where

f'

does not exist.

Why: The multiplication is only valid where $(x^2 - 1)^{1/3} \ne 0$ , which deletes exactly the points where $f'$ does not exist. Keep the power in the denominator and it keeps its column on the chart.

4. Reading a sign change at a number outside the domain as an extremum

1 to 2 marks, and a maximum with no value

What not to write

“ $g(x) = \frac{1}{(x - 1)^2}$ , $g'(x) = -\frac{2}{(x - 1)^3}$ goes from $+$ to $-$ at $1$ , so $g$ has a local maximum at $1$ .”

What to write

“ $g'$ changes sign at $1$ , but $1$ is not in the domain: no extremum. $g$ is increasing on $(-\infty, 1)$ and decreasing on $(1, \infty)$ .”

g' > 0

on the left of

1

and

g' < 0

on the right, yet the graph climbs to infinity on both sides: no maximum where

g

does not exist.

Why: The First Derivative Test is stated at a critical point, a number OF THE DOMAIN where $f$ is continuous. Asking for the maximum VALUE $g(1)$ shows at once that there is none.

5. Multiplying an inequality by a denominator of unknown sign

2 marks, half of the chart

What not to write

“ $\frac{x + 1}{x - 2} > 0$ , multiply by $x - 2$ : $x + 1 > 0$ , so $f$ is increasing on $(-1, \infty)$ .”

What to write

“ $\frac{x + 1}{x - 2} > 0$ when numerator and denominator have the same sign: $x < -1$ or $x > 2$ . $f$ is increasing on $(-\infty, -1]$ and on $(2, \infty)$ .”

Why: Multiplying by a negative number reverses an inequality, and $x - 2$ is negative for $x < 2$ . At $x = 0$ the faux line claims $f' > 0$ while $f'(0) = -\frac{1}{2}$ . Multiply only by a quantity of KNOWN sign, like $t^2 + 36$ .

6. Expanding a product with an exponential instead of factoring it

the whole question

What not to write

“ $f'(x) = 2xe^x - 5e^x + x^2e^x - 5xe^x + 5e^x = x^2e^x - 3xe^x$ , and I cannot solve $x^2e^x = 3xe^x$ .”

What to write

“ $f'(x) = e^x\left[(2x - 5) + (x^2 - 5x + 5)\right] = e^x(x^2 - 3x) = x(x - 3)e^x$ , and $e^x > 0$ .”

Why: In the product rule both terms contain $e^x$ : take it out BEFORE simplifying. The bracket then shrinks to a polynomial that factors, and $e^x$ leaves the sign discussion.

7. Leaving the endpoints of a closed interval out of the local extrema

1 mark per endpoint

What not to write

“On $[0, 2\pi]$ , the local extrema of $f(x) = \cos 2x + 2\sin x$ are at $\frac{\pi}{6}$ , $\frac{\pi}{2}$ , $\frac{5\pi}{6}$ and $\frac{3\pi}{2}$ .”

What to write

“And at the endpoints: $f$ increases right after $0$ , local minimum $f(0) = 1$ ; $f$ increases up to $2\pi$ , local maximum $f(2\pi) = 1$ .”

Why: Thomas defines local extrema at endpoints: $f(c)$ compared with the values of the domain near $c$ . When the question gives a closed interval, the two ends are on the list.

8. Refusing “increasing” because f' vanishes at one point

1 to 2 marks

What not to write

“ $h(x) = 2x - \sin 2x$ has $h'(\pi) = 0$ , so $h$ is not increasing on $[0, 2\pi]$ ; it has a critical point at $\pi$ , hence an extremum.”

What to write

“ $h'(x) = 4\sin^2 x > 0$ on $(0, \pi)$ and on $(\pi, 2\pi)$ , so $h$ is increasing on $[0, \pi]$ and on $[\pi, 2\pi]$ , hence on $[0, 2\pi]$ ; no extremum at $\pi$ .”

Why: A derivative that vanishes at isolated points does not stop the increase: apply the corollary on each side and join the intervals at the shared point.

Which method to choose

Which algebra gesture, by the form of f'

Look at the form of f' as you first write it, before any sign

If a polynomial with a common factor → take out the common factor, then factor the rest; check a quadratic's discriminant

Example: $12x^3 + 12x^2 + 12x = 12x(x^2 + x + 1)$
If a quotient $\frac{u}{v}$ → brackets around u v', expand, factor the numerator, keep the denominator factored

Example: $\frac{(x - 10)(x + 2)}{(x - 4)^2}$
If powers like $x^{5/3}$ and $x^{-1/3}$ , or a sum with a root → factor out the smallest exponent, or take the common denominator with the root

Example: $\frac{8}{3}x^{-1/3}(x^2 - 4)$ ; $\frac{3(4 - x)}{2\sqrt{6 - x}}$
If every term contains $e^x$ or $e^{g(x)}$ → factor the exponential first, it is positive and leaves the sign

Example: $e^x(x^2 - 3x) = x(x - 3)e^x$
If $\sin 2x$ , $\cos 2x$ with $\sin x$ or $\cos x$ → double-angle identity, then factor; never divide

Example: $-2\sin 2x + 2\cos x = 2\cos x(1 - 2\sin x)$
If $a\ln x$ or $\frac{a}{x}$ next to a polynomial → one fraction over x, and reject the roots outside x > 0

Example: $2x - \frac{8}{x} = \frac{2(x - 2)(x + 2)}{x}$ , $-2$ rejected

If no branch fits, test the derivative at two easy points before and after simplifying: two different values mean an algebra slip, and it is cheaper to find it now than on the chart.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Proving an inequality with the monotonicity corollary

When to use it: “Prove that $F(x) > G(x)$ for every $x > a$ ”, with no obvious algebraic route

1 Define the difference $d(x) = F(x) - G(x)$ on $[a, \infty)$ and state that it is continuous there, with the reason (sum of continuous functions).
2 Compute $d'$ and bring it to ONE fraction, factored: $\frac{1}{1 + x} - \frac{1}{(1 + x)^2} = \frac{x}{(1 + x)^2}$ .
3 State the sign of $d'$ on $(a, \infty)$ , factor by factor.
4 Name the corollary: $d$ is increasing on $[a, \infty)$ .
5 Compute the starting value $d(a)$ and conclude $d(x) > d(a)$ for $x > a$ .

Concluding sentence

“Since $d$ is continuous on $[0, \infty)$ and $d'(x) = \frac{x}{(1 + x)^2} > 0$ on $(0, \infty)$ , $d$ is increasing on $[0, \infty)$ ; hence for $x > 0$ , $d(x) > d(0) = 0$ , that is $\ln(1 + x) > \frac{x}{1 + x}$ .”

The trap: Skipping the value $d(a)$ : a positive derivative proves that $d$ goes up, not that it is positive.

Marking: Typically 1 mark for the difference and its continuity, 3 for d' simplified, 2 for its sign, 2 for the corollary named, 2 for the starting value and the conclusion.

Classifying the critical points on a closed interval

When to use it: “Find the local extreme values of $f$ on $[a, b]$ and where they occur”

1 Compute $f'$ and factor it completely, using identities before factoring.
2 List the critical points in $(a, b)$ : zeros of $f'$ AND points where $f'$ does not exist.
3 Sign chart on $[a, b]$ , one row per factor, one test value per interval.
4 Classify each interior critical point by the First Derivative Test, with its VALUE.
5 Classify the endpoints $a$ and $b$ by the sign of $f'$ next to them, with their values.

Concluding sentence

“Since $f'$ changes from positive to negative at $x = \frac{\pi}{6}$ , $f$ has a local maximum there by the First Derivative Test, with value $f\left(\frac{\pi}{6}\right) = \frac{3}{2}$ .”

The trap: Stopping at the interior points: the endpoints are part of the answer on a closed interval.

Marking: Typically 2 for f' factored, 2 for the full list of critical points, 3 for the chart, 3 for the classified extrema with values, endpoints included.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The factored form against the raw derivative

Evaluate the unsimplified derivative and your factored form at x = 0 and at one other easy point. They must agree; if not, the algebra slipped.

For $\frac{x^2 + 5x}{x - 4}$ : at $x = 0$ the right and the wrong numerators both give $-20$ , the faulty term vanishing there; at $x = 10$ the raw form gives $\frac{(25)(6) - 150}{36} = 0$ and the wrong $\frac{x^2 + 2x - 20}{(x - 4)^2}$ gives $\frac{25}{9}$ .
Maxima and minima alternate on each continuous piece

Between two local maxima of a continuous function on an interval there is a local minimum. Two maxima in a row on the same piece mean a sign error; across an asymptote the rule does not apply.

$\cos 2x + 2\sin x$ on $[0, 2\pi]$ : min, max, min, max, min, max, from $0$ to $2\pi$ , endpoints included.
The value matches the type

On one continuous piece, a local maximum value is above the neighbouring local minimum values. Compute the values and compare.

$x^5 - 5x^4 + 5x^3$ : max $1$ at $x = 1$ , min $-27$ at $x = 3$ , consistent; across the asymptote of $\frac{x^2 + 5x}{x - 4}$ , max $1$ below min $25$ is allowed.
Every column of the chart is in the domain, or has a double bar

Before concluding, check each column against the domain: a root of f' outside the domain is removed, a number outside the domain where the sign changes gets a double bar and no verdict.

$\frac{2(x - 2)(x + 2)}{x}$ on $x > 0$ : the column $-2$ must not appear at all.

The typical problem, taken apart

A product with an exponential: factor before you judge

Let $f(x) = (x^2 - 5x + 5)e^x$ . Find the intervals on which $f$ is increasing and decreasing, and its local extreme values, exactly.

Every conclusion must name the rule that gives it, as on a MATH 203 final.

The graph drawn after the study: a local maximum

5

x = 0

and a local minimum

-e^3

x = 3

, as the sign of

x(x - 3)

predicts.

Step 1

Product rule: $f'(x) = (2x - 5)e^x + (x^2 - 5x + 5)e^x$ .

Why

Write both terms before touching them. Every mark of the question depends on this line being complete.

Step 2

Factor $e^x$ first: $f'(x) = e^x\left[2x - 5 + x^2 - 5x + 5\right] = e^x(x^2 - 3x) = x(x - 3)e^x$ .

Why

Taking out $e^x$ BEFORE expanding leaves a small polynomial. Expanding first produces five terms and the classic slip on $-5x + 2x$ .

Step 3

$e^x > 0$ , so the sign of $f'$ is the sign of $x(x - 3)$ : $+$ on $(-\infty, 0)$ , $-$ on $(0, 3)$ , $+$ on $(3, \infty)$ . $f$ is increasing on $(-\infty, 0]$ , decreasing on $[0, 3]$ , increasing on $[3, \infty)$ .

Why

The always-positive factor leaves the sign discussion but stays in the derivative. The corollary is named for the intervals.

Step 4

First Derivative Test: $+$ to $-$ at $0$ , local maximum $f(0) = 5$ ; $-$ to $+$ at $3$ , local minimum $f(3) = (9 - 15 + 5)e^3 = -e^3$ .

Why

The question asks for VALUES: $5$ and $-e^3$ , with the points $0$ and $3$ . The exact form $-e^3$ is the expected answer, about $-20.1$ if the calculator checks it.

Step 5

Check: $f'(1) = (-3)e + (1)e = -2e$ from the raw form, and $1 \cdot (1 - 3)e = -2e$ from the factored one; and $5 > -e^3$ on the same continuous piece, as a maximum must be.

Why

Two ten-second checks: the factored form agrees with the raw derivative, and the values respect the order of the types.

The conclusion, written out

“ $f$ is increasing on $(-\infty, 0]$ and on $[3, \infty)$ and decreasing on $[0, 3]$ ; by the First Derivative Test it has a local maximum value $f(0) = 5$ and a local minimum value $f(3) = -e^3$ .”

The classic mistake on this problem: Expanding the product rule into five terms and dropping one, or dividing $x^2e^x = 3xe^x$ by $xe^x$ and losing the critical point $0$ .

Learn by heart

• Corollary: $f$ continuous on $[a, b]$ and $f' > 0$ on $(a, b)$ imply $f$ increasing on $[a, b]$ ; $f' < 0$ , decreasing.
• Critical point: in the DOMAIN, with $f'(c) = 0$ or $f'(c)$ undefined.
• First Derivative Test: $-$ to $+$ minimum, $+$ to $-$ maximum, no change nothing; $f$ continuous at $c$ .
• Read signs on ONE fraction, fully factored; even powers and never-zero factors do not change the sign.
• Never divide or multiply $f'(x) = 0$ by something that can vanish: factor it.
• Multiply an inequality only by a quantity of KNOWN sign.
• On a closed interval, the endpoints are local extrema too.

Frequently asked questions

How do I find where a function is increasing or decreasing in MATH 203?

Differentiate, then rewrite the derivative as one fraction with a completely factored numerator and denominator. List the numbers where it is zero or undefined, and make a sign chart with one row per factor. Where the derivative is positive the function is increasing, where it is negative it is decreasing, interval by interval, never across a point outside the domain.

What is the First Derivative Test for local extrema?

At a critical point where the function is continuous, look at the sign of the derivative just before and just after. From negative to positive gives a local minimum, from positive to negative a local maximum, and no change of sign means no local extremum. The test needs no second derivative, so it works at cusps and corners too.

Why does my sign chart give the wrong intervals when my derivative looks right?

Usually because the derivative was never fully factored, or a step lost information: a minus sign of the quotient rule applied to one term only, a cosine or a power of x divided away, or an inequality multiplied by a denominator that can be negative. Test the simplified derivative at two easy points against the raw one before building the chart.

Can an endpoint of an interval be a local maximum or minimum?

Yes, in Thomas's definition used in MATH 203. Compare the value at the endpoint with the values of the function at nearby points of the domain. If the function decreases right after the left endpoint, that endpoint is a local maximum; if it increases right after, a local minimum. On a closed interval both ends belong on the list of local extrema.

Is a function still increasing if its derivative is zero at one point?

Yes. If the derivative is positive on both sides of an isolated zero, the function is increasing on each side and the two intervals join at that point, so it is increasing on the whole interval. The cube function is the standard example: its derivative vanishes at the origin, and it still increases everywhere, with no extremum.

Practise it

Corrected exercises: Monotonic functions and the First Derivative Test, MATH 203 at Concordia

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Extreme values of functions Next sheet Concavity and curve sketching

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-monotonic-functions. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: monotonic functions and the First Derivative Test (MATH 203)

The essentials

One fraction, fully factored, then the chart

What the test needs, and what it concludes

The rules in table form

How each kind of factor behaves on the sign chart

The mistakes that cost marks

1. Applying the minus sign of the quotient rule to one term only

2. Dividing by cos x, or by x, to solve f'(x) = 0

3. Multiplying through by a fractional power and losing the cusp

4. Reading a sign change at a number outside the domain as an extremum

5. Multiplying an inequality by a denominator of unknown sign

6. Expanding a product with an exponential instead of factoring it

7. Leaving the endpoints of a closed interval out of the local extrema

8. Refusing “increasing” because f' vanishes at one point

Which method to choose

Which algebra gesture, by the form of f'

How the answer is expected to be written

Proving an inequality with the monotonicity corollary

Classifying the critical points on a closed interval

Check before you hand in

The typical problem, taken apart

A product with an exponential: factor before you judge

Learn by heart

Frequently asked questions

How do I find where a function is increasing or decreasing in MATH 203?

What is the First Derivative Test for local extrema?

Why does my sign chart give the wrong intervals when my derivative looks right?

Can an endpoint of an interval be a local maximum or minimum?

Is a function still increasing if its derivative is zero at one point?

Corrected exercises: Monotonic functions and the First Derivative Test, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?