Video Transcript
Given that 𝑓 of 𝑥 equals eight 𝑥 to the fourth power minus 16𝑥 squared plus five, determine the intervals on which 𝑓 is increasing or decreasing.
Let’s begin by recalling what we understand about an increasing or decreasing function. We say that a function is increasing when the first derivative of that function evaluated at a point is greater than zero. And it’s decreasing for values of 𝑥 such that the first derivative of 𝑓 of 𝑥 is less than zero. And so it follows that the first thing that we can do is find the first derivative of our function 𝑓 of 𝑥. Notice that 𝑓 of 𝑥 is a polynomial function. And we know that polynomials are differentiable over their entire domain. So we’re going to simply differentiate term by term. And, of course, to differentiate a power term of the form 𝑎𝑥 to the 𝑛th power, where 𝑎 and 𝑛 are real constants, we multiply the entire term by the exponent and then reduce that exponent by one.
So the first derivative of eight 𝑥 to the fourth power is four times eight 𝑥 to the power of four minus one, or eight 𝑥 cubed. Then the first derivative of 16𝑥 squared is two times 16𝑥. And of course the derivative of the sum or difference of two functions is equal to the sum or difference of the derivatives. So our first two terms are four times eight 𝑥 cubed minus two times 16𝑥. And, in fact, the derivative of any constant is zero. So the derivative of five is zero, which means that 𝑓 prime of 𝑥 simplifies to 32𝑥 cubed minus 32𝑥.
Our job, of course, is to find the values of 𝑥 such that this is less than and greater than zero. To do so, we could use an inequality solver on a calculator. Short of that though, we can consider the shape of the graph. Let’s look at that second method and just begin by finding where the first derivative is equal to zero. To do so, we factor the expression on the left-hand side. And we get 32𝑥 times 𝑥 squared minus one equals zero. And, of course, for the product of these two functions to be equal to zero, either one or either of the functions must themselves be equal to zero. So 32𝑥 could be equal to zero or 𝑥 squared minus one could be equal to zero. Now, dividing both sides of this first equation by 32, and we find that 𝑥 is equal to zero is a solution to our equation.
To solve our second equation though, we’ll begin by adding one to both sides. And then we take the square root of both sides of our equation, remembering, of course, to take both the positive and negative square root of one. And so we find that 𝑥 is equal to one or negative one as two further solutions to our equation. We now know that the graph of our derivative is a cubic which passes through the 𝑥-axis at zero, one, and negative one. It has a positive leading coefficient, so it looks a little bit like this.
We can use this graph to identify where the first derivative is greater than or less than zero. Well, it’s clearly greater than zero in these two locations and less than zero here. And so since the first derivative is greater than zero for values of 𝑥 greater than negative one and less than zero and values of 𝑥 greater than one. We say that the function 𝑓 is increasing on the open intervals from negative one to zero and one to ∞. Similarly, the first derivative is less than zero for values of 𝑥 less than negative one and between zero and one. So the function 𝑓 must be decreasing on the open interval from negative ∞ to negative one and zero to one.
And so we’ve identified the intervals of increase and decrease for our function. And, of course, remember that these intervals must be open as opposed to closed since 𝑓 prime of 𝑥 being equal to zero indicates a critical point.
