Question Video: Discussing the Monotonicity of Piecewise Functions Mathematics

Determine the intervals of increase and decrease of 𝑓(π‘₯) = βˆ’9π‘₯, if π‘₯ < 0 and 𝑓(π‘₯) = 9π‘₯, if π‘₯ β‰₯ 0.

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Video Transcript

Determine the intervals of increase and decrease of the piecewise-defined function 𝑓 of π‘₯ is equal to negative nine π‘₯ if π‘₯ is less than zero and 𝑓 of π‘₯ is equal to nine π‘₯ if π‘₯ is greater than or equal to zero.

In this question, we’re given a piecewise-defined function 𝑓 of π‘₯. And we’re asked to determine the intervals in increase and decrease of this function. And we can recall we say a function is increasing on an interval if taking a larger input value on the interval increases its output. And we say a function is decreasing on an interval if taking a larger input decreases its output on the interval. And this in fact gives us two ways of determining the intervals in increase and decrease of this function.

The first way we can do this is to sketch a graph of 𝑦 is equal to 𝑓 of π‘₯. We can do this by sketching each subfunction separately. Let’s start by sketching 𝑦 is equal to negative nine π‘₯ where π‘₯ must be less than zero. First, this is a straight line written in slope–intercept form. We can see its 𝑦-intercept is at zero and its slope is negative nine. But remember, we’re only sketching this subfunction for values of π‘₯ less than zero. So we need to have a hollow dot at this 𝑦-intercept. This gives us the following sketch.

We now need to sketch the graph of the second subfunction. That’s 𝑦 is equal to nine π‘₯ for values of π‘₯ greater than or equal to zero. Once again, this is a straight line in slope–intercept form. Its 𝑦-intercept is zero and its slope is nine. Since π‘₯ is equal to zero is included in this subfunction, we need to include the 𝑦-intercept which is at the origin. This gives us a sketch which looks like the following.

We can use this to determine the intervals of increase and decrease of our function. Let’s start with the intervals of increase. The intervals of increase of our function are the intervals where taking a larger input value increases the output. In the diagram, this will be any interval where our function is sloping upwards. Since this is a piecewise function made up of two straight lines, we can see this will only be true on the interval where π‘₯ is positive. So our function is only increasing when π‘₯ is positive. We want to write this as an interval. That’s the open interval from zero to ∞.

We can do exactly the same to determine the interval of decrease from the sketch. Once again, an interval of decrease is an interval where taking a larger input value of π‘₯ decreases the output value of the function. In the graph, this will be anywhere where our function is sloping downwards. And we can see on our diagram this is true when π‘₯ is negative. Therefore, from our sketch, we can see the interval of decrease of this function is the open interval from negative ∞ to zero. This is then enough to answer our question. We can say that our function 𝑓 of π‘₯ is increasing on the open interval from zero to ∞ and decreasing on the open interval from negative ∞ to zero.

We could stop here. However, there is a second method we can use to answer this question. We can notice that 𝑓 of π‘₯ is a piecewise linear function. In particular, since linear functions are differentiable, this means that 𝑓 of π‘₯ is a piecewise differentiable function. And we know that we can determine the intervals of increase and decrease of a differentiable function by using the sign of its derivative. In particular, if 𝑓 prime of π‘₯ is positive on an interval 𝐼, then we can say that 𝑓 is increasing on this interval 𝐼. And if 𝑓 prime of π‘₯ is negative on an interval 𝐼, then 𝑓 is decreasing on 𝐼.

And it’s worth noting we can see this on the graph we sketched. When the function has a negative slope, its outputs are decreasing. And when our function has a positive slope, its outputs are increasing. So we want to find an expression for 𝑓 prime of π‘₯. We can do this by differentiating each of the subfunctions separately since this is a piecewise differentiable function.

However, there is one small thing we need to note. We do need to consider what happens at the endpoints of our subdomains. In particular, in this case, we can know our function won’t be differentiable when π‘₯ is zero. And this is because as π‘₯ approaches zero from the left, its derivative is negative nine. However, as π‘₯ approaches zero from the right, its derivative is positive nine. And since the limit as π‘₯ approaches zero from the left and the limit as π‘₯ approaches zero from the right of this function are not equal, its limit as π‘₯ approaches zero does not exist. So we don’t include the endpoint of this interval.

𝑓 prime of π‘₯ is negative nine if π‘₯ is less than zero. And 𝑓 prime of π‘₯ is nine if π‘₯ is greater than zero. And we can then answer the question directly from this function. 𝑓 prime of π‘₯ is positive when π‘₯ is greater than zero, so it’s increasing when π‘₯ is greater than zero. And 𝑓 prime of π‘₯ is negative when π‘₯ is less than zero, so 𝑓 prime of π‘₯ is decreasing when π‘₯ is less than zero. And this gives us a second method of answering this question.

Therefore, we were able to show 𝑓 of π‘₯ is increasing on the open interval from zero to ∞ and is decreasing on the open interval from negative ∞ to zero.

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