# Video: Finding Intervals Where Absolute Value Functions Increase and Decrease

Determine the intervals over which the function π(π₯) = β|2π₯| + 28 is increasing and over which it is decreasing.

03:30

### Video Transcript

Determine the intervals over which the function π of π₯ equals the negative absolute value of two π₯ plus 28 is increasing and over which it is decreasing.

We begin by recalling how we generally calculate the intervals over which a function is increasing or decreasing. We say that a function is increasing when its first derivative is greater than zero. So, the interval over which a function is increasing will be the values of π₯ for which the first derivative is bigger than zero. Similarly, the interval over which a function is decreasing is calculated by finding the set of values of π₯ such that π prime of π₯ is less than zero.

Now, actually, that isnβt necessarily the quickest way to find the intervals of increase and decrease for our absolute-value function. But we will consider both methods. The first method is to sketch the graph of π of π₯ equals the negative absolute value of two π₯ plus 28. And in fact, sketching the graph actually helps us find the relevant functions to differentiate anyway. So, weβll need to do that either way. We begin by considering the graph of π¦ equals two π₯. Itβs a single straight line that passes through the origin and has a slope of two.

When we find the absolute value of two π₯, we take any values of two π₯ and essentially assure that theyβre positive. In terms of the graph then, any bits of the graph that lie below the π₯-axis are reflected in it, as shown. The graph of negative the absolute value of two π₯ isnβt simply a reflection of the entire graph of π¦ equals the absolute value of two π₯ in the π₯-axis. And then when we add 28, we move our graph or we translate it 28 units upwards.

And so, now, we have the graph of π¦ equals the negative absolute value of two π₯ plus 28. And actually, since this graph is essentially made up of two single straight lines, we can simply identify its intervals of increase and decrease. We see that the slope of the graph for values of π₯ less than zero is positive, so itβs increasing for values of π₯ less than zero. Conversely, it is decreasing for all values of π₯ greater than zero. Notice that the graph is sloping downwards. Its first derivative is negative. In interval notation, weβd say itβs decreasing over the open interval from zero to β and increasing over the open interval from negative β to zero.

But how could we be more stringent about this? How could we prove this using calculus? What will we need to do is consider the individual equations of each bit of our graph. We can say that the function is defined by two π₯ plus 28 for values of π₯ less than zero and negative two π₯ plus 28 for values of π₯ greater than zero. Essentially, we have a piecewise function. Letβs differentiate each part of our piecewise function to find π prime of π₯.

The derivative of two π₯ plus 28 is two. So, π prime of π₯ is two for values of π₯ less than zero. Similarly, π prime of π₯ is negative two for the part of the function defined by negative two π₯ plus 28. Notice that the first derivative is greater than zero always for values of π₯ less than zero. And itβs less than zero always for values of π₯ greater than zero.

Essentially, we have proved that the part of the function defined by π of π₯ equals two π₯ plus 28 is always increasing. And so, our function is increasing over the open interval from negative β to zero. Similarly, itβs decreasing for values of π₯ greater than zero. Thatβs the open interval from zero to β.