# Video: Finding the Intervals Where a Polynomial Function Increases and Decreases

Determine the intervals on which the function π(π₯) = π₯Β³ β 3π₯ β 2 is increasing or decreasing.

03:51

### Video Transcript

Determine the intervals on which the function π of π₯ equals π₯ cubed minus three π₯ minus two is increasing or decreasing.

First, we recall that a function π of π₯ is increasing when its slope, which is given by the first derivative π prime of π₯, is positive and decreasing when its slope, π prime of π₯, is negative. We will need to use differentiation to find an expression for π prime of π₯ and then find the intervals on which this is positive or negative. But we can also consider a sketch of what π of π₯ might look like.

Our function π of π₯ is a cubic function. And it has a positive leading coefficient. So our function π of π₯ looks a little bit like this. From our sketch, we can see that weβre expecting π of π₯ to be increasing on two intervals, those intervals which are now shaded in orange. And weβre expecting π of π₯ to be decreasing on one interval, the interval now shaded in pink. We can use this sketch to provide a sense check of what we find using differentiation.

First then, letβs find an expression for the first derivative π prime of π₯. And we can do this using the power rule of differentiation. We differentiate term by term. The derivative of π₯ cubed is three π₯ squared. The derivative of negative three π₯ is negative three. And the derivative of a constant, negative two, is zero. So we find that π prime of π₯ is equal to three π₯ squared minus three.

To determine the intervals on which the function π of π₯ is increasing, we need to consider where its first derivative is positive. So we can take our expression for the first derivative and form an inequality, three π₯ squared minus three is greater than zero. We can divide both sides of this inequality by three and then add one to each side to give the equivalent inequality π₯ squared is greater than one.

Now, we need to be a little bit careful here because this is a quadratic inequality. It isnβt correct to just square root each side and say that π₯ is greater than the square root of one or even that π₯ is greater than plus or minus the square root of one.

Instead, letβs consider a number line of values. For π₯ squared to be greater than one, π₯ can be positive or negative. But it must be the case that the absolute value of π₯ is greater than one, which means that π₯ must be either less than negative one or greater than positive one. We therefore have two distinct regions over which our function is increasing, values of π₯ less than negative one and values of π₯ greater than positive one. This is consistent with what we saw in our sketch.

To find where our function π of π₯ is decreasing, we can go through an almost identical process. But we just need to reverse the direction of the inequality. π of π₯ will be decreasing when three π₯ squared minus three β thatβs our first derivative β is less than zero, which leads to π₯ squared is less than one. For π₯ squared to be less than one, the absolute value of π₯ must also be less than one, which means that π₯ itself must be between negative one and one.

So we have a single interval over which our function is decreasing. And again, this is consistent with what we saw on our sketch of the graph of π of π₯. Notice that the values of negative one and one are not included in either interval. And this is because, at each of these values, π prime of π₯ is equal to zero. They are critical points of the function π of π₯.

By considering the sign of the first derivative of our function then, weβve seen that the function is increasing on the open intervals negative β to negative one and one to β and decreasing on the open interval negative one to one.