Increasing & Decreasing Functions Revision Notes for Leaving Cert Mathematics

Increasing & Decreasing Functions

Consider the curve $x^3-2x^2+2$ . You should notice that in some regions, the curve is increasing and in others the curve is decreasing :

The portion of the curve coloured red represents the portion of the curve where it is decreasing (i.e. going down from left to right). As a result, all the slopes of the tangents at these points, will have a negative slope.

On the contrary, tangents touching points on the black regions will have positive slopes.

Because we know the nature of the slopes, we can use the first derivative to determine the portion of the graph for which the graph is either increasing or decreasing.

Example

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Find the range of $x$ values for which the graph is decreasing for $f(x)=3x^2-2x^3$ .

The region for which the function is decreasing means that the slope has to be negative.

f'(x)<0

6x-6x^2<0

x(6-6x)<0

x<0

(6-6x)<0 \implies 1<x

0>x>1

Example

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A function is defined on $\mathbb{R} \backslash \{1\}$ by $f(x)=\frac{x}{1-x}$ . Show that $f$ is an increasing function on $\mathbb{R} \backslash \{1\}$ .

We need to show that for all values in $\mathbb{R} \backslash \{1\}$ , the slopes of the tangents are positive. First, differentiate the function $f$ .

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

$f'(x)$ is positive for all values within the range. $(1-x)^2$ is a positive number since anything squared is positive. $\frac{1}{(1-x)^2}$ is positive since a positive divided by a positive, is also positive.

Increasing & Decreasing Functions (Leaving Cert Mathematics): Revision Notes