# Strictly Increasing & Decreasing Graphs

## Strictly Increasing Graphs

A graph $y = f(x)$ is said to be **strictly increasing** if

- b > a implies that f(b) > f(a) … for the entire domain of f(x)

This means that for all points in the domain, if we take a step of any size to the right, the y-value must have increased.

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This can include **stationary points**.

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#### Example 1

In the graph shown here:

… … $y = f(x)$ is a strictly increasing graph over the entire domain, $x \in R$.

… … … including at x = 1 (stationary point)

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… … $y = g(x)$ is a strictly increasing graph in the domain: $x \in \big( -\infty, \; 2 \big]$

… … … including at x = 2 (stationary point)

**Note**

- If a graph is strictly increasing (or decreasing) then its inverse function is defined.

- There is no requirement that the graph be differentiable or continuous.

Strictly increasing (and decreasing) graphs are described as **one-to-one** functions.

- for any
**x**value, there is only one**y**value and - for any
**y**value, there is only one**x**value}

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#### Example 2

In the graph shown here:

… … $y = f(x)$ is a strictly increasing graph over the entire domain $x \in R$

…. … … including at x = 1 where it is not differentiable (not smooth)

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… … $y = g(x)$ is a strictly increasing graph over the entire domain $x \in R$

… … … including at x = 2 where it is not differentiable (not continuous)

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## Strictly Decreasing Graphs

In the same way, a graph $y = f(x)$ is said to be **strictly decreasing** if

- b > a implies that f(b) < f(a) … for the entire domain of f(x)

- This means that if we take a step of any size to the right, the y value must have decreased.

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#### Example 3

Identify the sections where the graph of $y = \big( x^2 - 1\big)^2$ is strictly decreasing.

**Solution**

The graph of $y = \big( x^2 - 1\big)^2$ is strictly decreasing in two sections:

… … $\big\{ x:x \leqslant -1 \big\}$

and

… … $\big\{ x: 0 \leqslant x \leqslant 1 \big\}$

$Note:$ The graph is **not** strictly decreasing in the combined domain:

… … $\big\{ x:x \leqslant -1 \big\} \cup \big\{ x: 0 \leqslant x \leqslant 1 \big\}$

because in this domain, there are pairs of points where a step to the right does **not** result in a decrease.

- eg between x = –1 and x = 0
- or between x = –1.2 and x = 0.2

- Also: it is not a one-to-one graph in this domain.

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We could describe a combined domain which does produce a strictly decreasing graph.

For example: $\big\{ x:x < -\sqrt{2} \big\} \cup \big\{ x: 0 \leqslant x \leqslant 1 \big\}$

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