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

Determine the intervals on which the function 𝑓(π‘₯) = π‘₯Β³ βˆ’ 3π‘₯ βˆ’ 2 is increasing or decreasing.

03:51

Video Transcript

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

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

Our function 𝑓 of π‘₯ is a cubic function. And it has a positive leading coefficient. So our function 𝑓 of π‘₯ looks a little bit like this. From our sketch, we can see that we’re expecting 𝑓 of π‘₯ to be increasing on two intervals, those intervals which are now shaded in orange. And we’re expecting 𝑓 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, let’s find an expression for the first derivative 𝑓 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 𝑓 prime of π‘₯ is equal to three π‘₯ squared minus three.

To determine the intervals on which the function 𝑓 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 isn’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, let’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 𝑓 of π‘₯ is decreasing, we can go through an almost identical process. But we just need to reverse the direction of the inequality. 𝑓 of π‘₯ will be decreasing when three π‘₯ squared minus three β€” that’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 𝑓 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, 𝑓 prime of π‘₯ is equal to zero. They are critical points of the function 𝑓 of π‘₯.

By considering the sign of the first derivative of our function then, we’ve seen that the function is increasing on the open intervals negative ∞ to negative one and one to ∞ and decreasing on the open interval negative one to one.

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