Increasing and Decreasing Functions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Increasing means rising from left to right

An increasing function gives a larger output when the input increases. A decreasing function gives a smaller output. This describes a direction, not whether the curve is above or below the axis.

The derivative gives the tangent gradient. Positive gradient means rising; negative gradient means falling. An interval is a continuous range of input values.

\begin{aligned}f'(x)>0&:\ \text{increasing}\\f'(x)<0&:\ \text{decreasing}\end{aligned}

Point position x0

Blue curve: y = x³ − 3x. Dashed green line: tangent.

Curve height f(x): 0
Gradient f′(x): -3

Decreasing

Move through $x=-1$ and $x=1$ . The gradient reaches zero at each horizontal tangent, then changes sign.

Find where the derivative has each sign

Worked example

Find where f(x) = x³ − 3x increases and decreases.

\begin{aligned}f'(x)&=3x^2-3\\&=3(x-1)(x+1)\end{aligned}

The zeros $-1,1$ split the number line into three intervals. A continuous derivative cannot change sign inside one of these intervals without another zero. Test one value in each:

$x=-2$ : both brackets are negative, so their product is positive.
$x=0$ : the brackets have opposite signs, so the product is negative.
$x=2$ : both brackets are positive.

Answer

\begin{gathered}x<-1\text{ or }x>1\\\text{(increasing)}\\-1<x<1\\\text{(decreasing)}\end{gathered}

Use “or” for the two separate increasing intervals. The strict inequalities above describe where the gradient is non-zero.

A zero derivative at one point does not by itself stop a function being increasing: $x^3$ increases across zero, although its gradient is zero there. Check the behaviour on both sides.

Your turnPractice

Find where $g(x)=6x-x^2$ decreases. Explain why a positive value of $g(x)$ is not enough to decide.

Show worked answer

\begin{gathered}g'(x)=6-2x<0\\x>3\end{gathered}

The derivative controls direction. For example, $g(4)=8>0$ but $g'(4)=-2<0$ .

Keep the original domain

Split at excluded values as well as derivative zeros. Never join intervals across a point where the function is undefined.

Worked example

Show that f(x) = 5/(2x − 3) decreases for x > 3/2.

f'(x)=-\frac{10}{(2x-3)^2}

The squared denominator is positive throughout the given domain. The numerator is negative, so the derivative is always negative. No equation needs solving.

Your turn9709/12 Feb/Mar 2020 Q1 [3 marks]

For $f(x)=\dfrac1{3x+2}+x^2$ , $x<-1$ , determine whether the function is increasing, decreasing or neither.

Show worked answer

f'(x)=-\frac3{(3x+2)^2}+2x

Both terms are negative when $x<-1$ . Therefore $f'(x)<0$ throughout the domain: the function is decreasing.

Increasing means rising from left to right

An increasing function gives a larger output when the input increases. A decreasing function gives a smaller output. This describes a direction, not whether the curve is above or below the axis.

The derivative gives the tangent gradient. Positive gradient means rising; negative gradient means falling. An interval is a continuous range of input values.

\begin{aligned}f'(x)>0&:\ \text{increasing}\\f'(x)<0&:\ \text{decreasing}\end{aligned}

Point position x0

Blue curve: y = x³ − 3x. Dashed green line: tangent.

Curve height f(x): 0
Gradient f′(x): -3

Decreasing

Move through $x=-1$ and $x=1$ . The gradient reaches zero at each horizontal tangent, then changes sign.

Find where the derivative has each sign

Worked example

Find where f(x) = x³ − 3x increases and decreases.

\begin{aligned}f'(x)&=3x^2-3\\&=3(x-1)(x+1)\end{aligned}

The zeros $-1,1$ split the number line into three intervals. A continuous derivative cannot change sign inside one of these intervals without another zero. Test one value in each:

$x=-2$ : both brackets are negative, so their product is positive.
$x=0$ : the brackets have opposite signs, so the product is negative.
$x=2$ : both brackets are positive.

Answer

\begin{gathered}x<-1\text{ or }x>1\\\text{(increasing)}\\-1<x<1\\\text{(decreasing)}\end{gathered}

Use “or” for the two separate increasing intervals. The strict inequalities above describe where the gradient is non-zero.

A zero derivative at one point does not by itself stop a function being increasing: $x^3$ increases across zero, although its gradient is zero there. Check the behaviour on both sides.

Your turnPractice

Find where $g(x)=6x-x^2$ decreases. Explain why a positive value of $g(x)$ is not enough to decide.

Show worked answer

\begin{gathered}g'(x)=6-2x<0\\x>3\end{gathered}

The derivative controls direction. For example, $g(4)=8>0$ but $g'(4)=-2<0$ .

Keep the original domain

Split at excluded values as well as derivative zeros. Never join intervals across a point where the function is undefined.

Worked example

Show that f(x) = 5/(2x − 3) decreases for x > 3/2.

f'(x)=-\frac{10}{(2x-3)^2}

The squared denominator is positive throughout the given domain. The numerator is negative, so the derivative is always negative. No equation needs solving.

Your turn9709/12 Feb/Mar 2020 Q1 [3 marks]

For $f(x)=\dfrac1{3x+2}+x^2$ , $x<-1$ , determine whether the function is increasing, decreasing or neither.

Show worked answer

f'(x)=-\frac3{(3x+2)^2}+2x

Both terms are negative when $x<-1$ . Therefore $f'(x)<0$ throughout the domain: the function is decreasing.