# Question Video: Determining the Monotonicity of a Function over an Interval

Which of the following is true of the function 𝑓(𝑥) = 𝑥² − 1, where 𝑥 ≤ 0? [A] 𝑓(𝑥) is increasing on the interval (0, ∞). [B] 𝑓(𝑥) is decreasing on the interval (0, ∞), and increasing on the interval (−∞, 0). [C] 𝑓(𝑥) is decreasing on the interval (−∞, 0). [D] 𝑓(𝑥) is increasing on the interval (0, ∞) and decreasing on the interval (−∞, 0). [E] 𝑓(𝑥) is increasing on the interval (−∞, 0).

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### Video Transcript

Which of the following is true of the function 𝑓 of 𝑥 equals 𝑥 squared minus one, where 𝑥 is less than or equal to zero?

Well we have five options to choose from here. The first one option A is that 𝑓 of 𝑥 is increasing on the interval zero, ∞, where neither end point is included in the interval; it’s an open interval. Option B is that 𝑓 of 𝑥 is decreasing on the interval zero, ∞ and increasing on the interval negative ∞, zero. Option C is that 𝑓 of 𝑥 is decreasing on the interval negative ∞, zero. Option D is that 𝑓 of 𝑥 is increasing on the interval zero, ∞ and decreasing on the interval negative ∞, zero. And finally, option E is that 𝑓 of 𝑥 is increasing on the interval negative ∞, zero.

All of these options have something to do with whether function 𝑓 of 𝑥 is increasing or decreasing, and so this suggests that we should draw a graph. The function we have 𝑓 of 𝑥 equals 𝑥 squared minus one is a quadratic function with constant term negative one. And so the 𝑦-intercept on the graph must be at 𝑦 equals negative one. What else is it helpful to know to draw this graph accurately? Well, it’s a quadratic function. So let’s find its root.

We can recognize 𝑥 squared minus one as the difference of two squares. It’s 𝑥 plus one times 𝑥 minus one. And so its roots are negative one and one. So now we have three points on our graph, and we can draw a smooth curve — a parabola — going through those points. And we noticed that it is an upward facing curve, which makes sense given that the coefficient of 𝑥 squared is one, which is positive.

Okay, so are we ready to go through the options one by one yet? Well, not quite; there’s also this bit here. It says “where 𝑥 is less than or equal to zero,” which defined the domain of our function. We therefore need to get rid of the part of the graph which is to the right of the 𝑦-axis.

Okay, now we’re ready. Let’s go through the options one by one. Is 𝑓 of 𝑥 increasing on the interval zero, ∞? Well, either looking at the graph or the definition in the question, we see that actually 𝑓 of 𝑥 isn’t even defined on this interval. So it can’t be increasing there. So this isn’t our correct option.

For a similar reason, we can rule out option B which says that 𝑓 of 𝑥 is decreasing on the interval zero, ∞. Well, actually 𝑓 of 𝑥 isn’t defined on that interval.

How about option C: 𝑓 of 𝑥 is decreasing on the interval negative ∞, zero? Well, 𝑓 of 𝑥 is at least defined on this interval. And if we go along the 𝑥-axis in the positive direction, we can see that the values of 𝑓 of 𝑥 are getting smaller as 𝑥 increases. For example, the value of 𝑓 of 𝑥 for this value of 𝑥 is less than the value of 𝑥 for a smaller value of 𝑥. As 𝑥 increases, 𝑓 of 𝑥 decreases and this happens all the way to the end point, zero. This therefore is our answer.

Let’s have a look at the other two options that are left to make sure we can see why those are false. Option D says that 𝑓 of 𝑥 is increasing on the interval zero, ∞. But of course, we’ve said that 𝑓 of 𝑥 is not defined there; it’s not part of the domain and so 𝑓 of 𝑥 can’t be increasing there. So that’s not true.

And finally, option E that 𝑓 of 𝑥 is increasing on the interval negative ∞, zero. Well, 𝑓 of 𝑥 is at least defined on this interval, but it’s decreasing on this interval as we saw earlier because as 𝑥 increases, the values of 𝑓 of 𝑥 decrease. For it’s be increasing as 𝑥 increases, the values of 𝑓 of 𝑥 would have to increase. And so this option E is also false.