Determine the intervals on which the function 𝑓 of 𝑥 equals 𝑥 cubed minus three 𝑥 minus two is increasing or decreasing.
First, we recall that a function 𝑓 of 𝑥 is increasing when its slope, which is given by the first derivative 𝑓 prime of 𝑥, is positive and decreasing when its slope, 𝑓 prime of 𝑥, is negative. We will need to use differentiation to find an expression for 𝑓 prime of 𝑥 and then find the intervals on which this is positive or negative. But we can also consider a sketch of what 𝑓 of 𝑥 might look like.
Our function 𝑓 of 𝑥 is a cubic function. And it has a positive leading coefficient. So our function 𝑓 of 𝑥 looks a little bit like this. From our sketch, we can see that we’re expecting 𝑓 of 𝑥 to be increasing on two intervals, those intervals which are now shaded in orange. And we’re expecting 𝑓 of 𝑥 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 𝑥. And we can do this using the power rule of differentiation. We differentiate term by term. The derivative of 𝑥 cubed is three 𝑥 squared. The derivative of negative three 𝑥 is negative three. And the derivative of a constant, negative two, is zero. So we find that 𝑓 prime of 𝑥 is equal to three 𝑥 squared minus three.
To determine the intervals on which the function 𝑓 of 𝑥 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 𝑥 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 𝑥 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 𝑥 is greater than the square root of one or even that 𝑥 is greater than plus or minus the square root of one.
Instead, let’s consider a number line of values. For 𝑥 squared to be greater than one, 𝑥 can be positive or negative. But it must be the case that the absolute value of 𝑥 is greater than one, which means that 𝑥 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 𝑥 less than negative one and values of 𝑥 greater than positive one. This is consistent with what we saw in our sketch.
To find where our function 𝑓 of 𝑥 is decreasing, we can go through an almost identical process. But we just need to reverse the direction of the inequality. 𝑓 of 𝑥 will be decreasing when three 𝑥 squared minus three — that’s our first derivative — is less than zero, which leads to 𝑥 squared is less than one. For 𝑥 squared to be less than one, the absolute value of 𝑥 must also be less than one, which means that 𝑥 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 𝑥. 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 𝑥 is equal to zero. They are critical points of the function 𝑓 of 𝑥.
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.