5.2 Increasing and decreasing functions (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Increasing and decreasing functions

A function is increasing on an interval if its graph rises from left to right there: as $x$ increases, $f(x)$ increases. It is decreasing if its graph falls from left to right.

The derivative tells you which: $f'(x)>0\ \Rightarrow\ f \text{ increasing}, \qquad f'(x)<0\ \Rightarrow\ f \text{ decreasing}.$ This is because $f'(x)$ is the gradient of the tangent: a positive gradient means the curve is going up.

Key termsincreasingdecreasing

Exam tip

Always answer with intervals of $x$ , not $y$ -values: 'decreasing for $-2<x<2$ '.

Section 2

Finding intervals where a function increases or decreases

Differentiate to find $f'(x)$ .
Solve $f'(x)=0$ to find the critical values where the sign may change.
Decide the sign of $f'(x)$ on each interval between them, using test values or the shape of the graph of $f'$ .
State the intervals.

Example: $f(x)=x^{3}-12x+1$ gives $f'(x)=3x^{2}-12=3(x-2)(x+2)$ . This positive quadratic is negative between its roots, so $f$ is decreasing for $-2<x<2$ and increasing for $x<-2$ or $x>2$ .

Key termscritical values

Common mistake

Solving $3x^{2}-12=0$ as $x=\pm4$ : you must divide by 3 and then take the square root, $x^{2}=4$ , $x=\pm2$ .

Exam tip

For a quadratic $f'(x)$ with positive leading coefficient: negative between the roots, positive outside them.

Section 3

Graphical interpretation of the sign of f′

If you are told about the graph of $y=f'(x)$ , translate it into the behaviour of $f$ :

where the graph of $f'$ is above the $x$ -axis, $f'(x)>0$ and $f$ is increasing;
where it is below the $x$ -axis, $f'(x)<0$ and $f$ is decreasing;
where it meets the $x$ -axis, $f'(x)=0$ and the tangent to $f$ is horizontal.

If $f'$ crosses the axis the direction of $f$ changes (a turning point). If $f'$ only touches the axis, $f$ keeps the same direction on both sides: the tangent is horizontal for a moment but $f$ does not turn.

Key termshorizontal tangentchange of sign

Common mistake

Confusing $f'(3)=0$ with $f(3)=0$ . The derivative being zero says the gradient is zero, not that the graph meets the $x$ -axis.

Section 4

Functions that are increasing everywhere

To show $f'(x)>0$ for all $x$ when $f'$ is a quadratic with positive leading coefficient, show it has no real roots: its discriminant is negative.

Example: $f(x)=x^{3}-3x^{2}+kx$ has $f'(x)=3x^{2}-6x+k$ . Then $f'(x)>0$ for all $x$ when $36-12k<0$ , i.e. $k>3$ .

Completing the square also works: $3x^{2}-6x+k=3(x-1)^{2}+k-3$ , which is always positive when $k>3$ .

Key termsdiscriminant

Exam tip

A function whose derivative is always positive is one-to-one, so it has an inverse. This links 5.2 to Topic 2.

Section 5

Using increasing and decreasing in context

In a model, $f'(t)<0$ means the quantity is falling. Interpret a derivative value with its sign, size and units: $V'(5)=-12$ thousand m $^{3}$ per month means the reservoir is losing water at 12 000 m $^{3}$ per month at that moment.

To find the greatest value of a quantity over a closed interval, use the increasing/decreasing intervals to list the candidates (points where $f$ stops increasing, and the endpoints), then compare their values.

Example

For $V(t)=t^{3}-15t^{2}+63t+200$ , $0\le t\le12$ : $V$ rises until $t=3$ ( $V=281$ ), falls until $t=7$ , then rises to $V(12)=524$ . The greatest volume is at $t=12$ .

Must know

$f'(x)>0$ : increasing; $f'(x)<0$ : decreasing; $f'(x)=0$ : horizontal tangent.
Solve $f'(x)=0$ , then test signs on each interval.
A sign change in $f'$ means $f$ turns; touching the axis without crossing means it does not.
Increasing for all $x$ : show $f'(x)>0$ always, e.g. negative discriminant.
State answers as intervals of $x$ and interpret in context with units.

That's the notes covered.

Carry on to the next subtopic.