IB Mathematics Analysis and Approaches Topic SL…

The Sign Of The Derivative Tells You Which Way A Graph Is Going

Once you can find a derivative, you get a powerful free gift: without plotting a single point, the sign of $f'(x)$ tells you whether the graph is climbing, falling, or momentarily flat.
This is the bridge between the algebra of derivatives and the shape of a curve.
It is the first step toward finding maximum and minimum points, which is where calculus earns its keep.

Increasing And Decreasing, Defined By Direction Of Travel

Read a graph left to right, in the direction of increasing $x$.
A function is increasing on an interval if its graph rises as you move right: larger $x$ gives larger $f(x)$.
A function is decreasing on an interval if its graph falls as you move right: larger $x$ gives smaller $f(x)$.
A function need not be one or the other everywhere.
Most curves increase on some intervals and decrease on others, switching direction at turning points.

Positive Gradient Climbs, Negative Gradient Falls, Zero Gradient Pauses

The gradient of the tangent is the rate at which the graph rises.
So the sign of $f'(x)$ reads directly as direction.
1. $f'(x) > 0$: the tangent slopes upward, so the function is increasing there.
2. $f'(x) < 0$: the tangent slopes downward, so the function is decreasing there.
3. $f'(x) = 0$: the tangent is horizontal, so the function is momentarily stationary (a candidate turning point).

Sign of f'(x)	Tangent	Behaviour of f
Positive	Slopes up	Increasing
Zero	Horizontal	Stationary
Negative	Slopes down	Decreasing

Note

The Method In One Line

To find where a function increases or decreases, you solve the inequality $f'(x) > 0$ or $f'(x) < 0$.
The values of $x$ where $f'(x) = 0$ are the boundaries between those intervals.

Finding The Intervals: A Worked Method

The routine is always the same: differentiate, find where the derivative is zero, then test the sign on each side.

Differentiate to get $f'(x)$.
Solve $f'(x) = 0$ to locate the stationary points.
These points split the number line into intervals; test the sign of $f'(x)$ in each.

Example

To find where $f(x) = x^3 - 3x$ is increasing and where it is decreasing:

Differentiate: $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$.
Solve $f'(x) = 0$: $x = -1$ and $x = 1$.
For $x < -1$ (try $x = -2$): $f'(-2) = 9 > 0$, so increasing.
For $-1 < x < 1$ (try $x = 0$): $f'(0) = -3 < 0$, so decreasing.
For $x > 1$ (try $x = 2$): $f'(2) = 9 > 0$, so increasing.
Thus, it is increasing for $x < -1$ and $x > 1$, decreasing for $-1 < x < 1$.

Example question

Find the intervals on which $f(x)=x^3-12x$ is increasing.

Solution

Differentiate: $$f'(x)=3x^2-12=3(x-2)(x+2)$$
Solve $f'(x)=0$: $x=-2$ and $x=2$.
$f'(x)>0$ for $x<-2$ and for $x>2$, so $f$ is increasing on those intervals.

Example question

Find the interval on which $f(x) = x^2 - 6x + 5$ is decreasing.

Solution

Differentiate: $$f'(x) = 2x - 6$$
Set $f'(x) < 0$: $2x - 6 < 0 \Rightarrow x < 3$.
So $f$ is decreasing for $x < 3$ (and increasing for $x > 3$).

Example question

The derivative of a function is $f'(x)=(x-1)(x-4)$. State the values of $x$ for which $f$ is increasing.

Solution

$f$ increases where $f'(x)>0$.
$(x-1)(x-4)>0$ when $x<1$ or $x>4$.
So $f$ is increasing for $x<1$ and for $x>4$.

Exam technique

This subtopic is a method, so nothing here is quoted from the booklet directly.
There is no formula-booklet entry for the increasing or decreasing test; you must know that $f'(x) > 0$ means increasing and $f'(x) < 0$ means decreasing.
The derivative you compute along the way does come from the booklet: the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ sits in the SL 5.3 section.
Command terms to expect: "find the values of $x$ for which $f$ is increasing" wants the interval, so solve the inequality, do not just list the stationary points.

Active recall

State the connection between the sign of $f'(x)$ and whether $f$ is increasing or decreasing.
What is true about $f'(x)$ at a stationary point?
Describe the three-step method for finding the intervals on which a function increases.
For $f(x) = x^3 - 3x$, on which intervals is $f$ increasing?
Why can a function be increasing even when all its output values are negative?

The Sign Of The Derivative Tells You Which Way A Graph Is Going

Once you can find a derivative, you get a powerful free gift: without plotting a single point, the sign of $f'(x)$ tells you whether the graph is climbing, falling, or momentarily flat.
This is the bridge between the algebra of derivatives and the shape of a curve.
It is the first step toward finding maximum and minimum points, which is where calculus earns its keep.

Increasing And Decreasing, Defined By Direction Of Travel

Read a graph left to right, in the direction of increasing $x$.
A function is increasing on an interval if its graph rises as you move right: larger $x$ gives larger $f(x)$.
A function is decreasing on an interval if its graph falls as you move right: larger $x$ gives smaller $f(x)$.
A function need not be one or the other everywhere.
Most curves increase on some intervals and decrease on others, switching direction at turning points.

Positive Gradient Climbs, Negative Gradient Falls, Zero Gradient Pauses

The gradient of the tangent is the rate at which the graph rises.
So the sign of $f'(x)$ reads directly as direction.
1. $f'(x) > 0$: the tangent slopes upward, so the function is increasing there.
2. $f'(x) < 0$: the tangent slopes downward, so the function is decreasing there.
3. $f'(x) = 0$: the tangent is horizontal, so the function is momentarily stationary (a candidate turning point).

Sign of f'(x)	Tangent	Behaviour of f
Positive	Slopes up	Increasing
Zero	Horizontal	Stationary
Negative	Slopes down	Decreasing

Note

The Method In One Line

To find where a function increases or decreases, you solve the inequality $f'(x) > 0$ or $f'(x) < 0$.
The values of $x$ where $f'(x) = 0$ are the boundaries between those intervals.

Finding The Intervals: A Worked Method

The routine is always the same: differentiate, find where the derivative is zero, then test the sign on each side.

Differentiate to get $f'(x)$.
Solve $f'(x) = 0$ to locate the stationary points.
These points split the number line into intervals; test the sign of $f'(x)$ in each.

Example

To find where $f(x) = x^3 - 3x$ is increasing and where it is decreasing:

Differentiate: $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$.
Solve $f'(x) = 0$: $x = -1$ and $x = 1$.
For $x < -1$ (try $x = -2$): $f'(-2) = 9 > 0$, so increasing.
For $-1 < x < 1$ (try $x = 0$): $f'(0) = -3 < 0$, so decreasing.
For $x > 1$ (try $x = 2$): $f'(2) = 9 > 0$, so increasing.
Thus, it is increasing for $x < -1$ and $x > 1$, decreasing for $-1 < x < 1$.

Example question

Find the intervals on which $f(x)=x^3-12x$ is increasing.

Solution

Differentiate: $$f'(x)=3x^2-12=3(x-2)(x+2)$$
Solve $f'(x)=0$: $x=-2$ and $x=2$.
$f'(x)>0$ for $x<-2$ and for $x>2$, so $f$ is increasing on those intervals.

Example question

Find the interval on which $f(x) = x^2 - 6x + 5$ is decreasing.

Solution

Differentiate: $$f'(x) = 2x - 6$$
Set $f'(x) < 0$: $2x - 6 < 0 \Rightarrow x < 3$.
So $f$ is decreasing for $x < 3$ (and increasing for $x > 3$).

Example question

The derivative of a function is $f'(x)=(x-1)(x-4)$. State the values of $x$ for which $f$ is increasing.

Solution

$f$ increases where $f'(x)>0$.
$(x-1)(x-4)>0$ when $x<1$ or $x>4$.
So $f$ is increasing for $x<1$ and for $x>4$.

Exam technique

This subtopic is a method, so nothing here is quoted from the booklet directly.
There is no formula-booklet entry for the increasing or decreasing test; you must know that $f'(x) > 0$ means increasing and $f'(x) < 0$ means decreasing.
The derivative you compute along the way does come from the booklet: the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ sits in the SL 5.3 section.
Command terms to expect: "find the values of $x$ for which $f$ is increasing" wants the interval, so solve the inequality, do not just list the stationary points.

Active recall

State the connection between the sign of $f'(x)$ and whether $f$ is increasing or decreasing.
What is true about $f'(x)$ at a stationary point?
Describe the three-step method for finding the intervals on which a function increases.
For $f(x) = x^3 - 3x$, on which intervals is $f$ increasing?
Why can a function be increasing even when all its output values are negative?

IB Mathematics Analysis and Approaches SL 5.2—increasing and Decreasing Functions Notes

IB Mathematics Analysis and Approaches SL 5.2—increasing and Decreasing Functions Notes

The Sign Of The Derivative Tells You Which Way A Graph Is Going

Increasing And Decreasing, Defined By Direction Of Travel

Positive Gradient Climbs, Negative Gradient Falls, Zero Gradient Pauses

Finding The Intervals: A Worked Method

The Sign Of The Derivative Tells You Which Way A Graph Is Going

Increasing And Decreasing, Defined By Direction Of Travel

Positive Gradient Climbs, Negative Gradient Falls, Zero Gradient Pauses

Finding The Intervals: A Worked Method

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 5.2—increasing and Decreasing Functions Notes

IB Mathematics Analysis and Approaches SL 5.2—increasing and Decreasing Functions Notes

The Sign Of The Derivative Tells You Which Way A Graph Is Going

Increasing And Decreasing, Defined By Direction Of Travel

Positive Gradient Climbs, Negative Gradient Falls, Zero Gradient Pauses

Finding The Intervals: A Worked Method

The Sign Of The Derivative Tells You Which Way A Graph Is Going

Increasing And Decreasing, Defined By Direction Of Travel

Positive Gradient Climbs, Negative Gradient Falls, Zero Gradient Pauses

Finding The Intervals: A Worked Method