### Video Transcript

Determine the intervals on which the function π of π₯ is equal to π₯ to the fourth power over two minus four π₯ squared plus two is increasing or decreasing.

The question gives us the function π of π₯, which is a polynomial, and it wants us to find the intervals on which this function is increasing or decreasing. And we recall, for a differential function π, we say that π is decreasing when its derivative is less than zero. And we say that π is increasing when its derivative is greater than zero. Since weβre given the function π of π₯, which is a polynomial, itβs differentiable for all real numbers. So, we just need to find our derivative function π prime of π₯.

We can do this by using the power rule for differentiation, which tells us, for any constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to π times π times π₯ to the power of π minus one. Applying this to each term in our function π of π₯ gives us four times π₯ to the power of four minus one over two minus two times four times π₯ to the power of two minus one. And the derivative of the constant two is just equal to zero. We can then simplify this to get two π₯ cubed minus eight π₯. We want to find the values of π₯ where this is negative and the values of π₯ where this is positive. Since our derivative function π prime of π₯ is a cubic polynomial, weβre going to do this by sketching a graph of π prime of π₯.

Letβs start by finding the π₯-intercepts of our polynomial. We see that both terms share a factor of π₯, and both terms share a factor of two. So, we can take out a factor of two π₯. This gives us two π₯ multiplied by π₯ squared minus four. And we can then see that π₯ squared minus four is a difference between squares. So, we can factor this to give us two π₯ multiplied by π₯ minus two multiplied by π₯ plus two. To find the π₯-intercepts, we want to find the values where this is equal to zero. And we know if a product of three numbers is equal to zero, then one of the factors must be equal to zero. Since the π₯-intercepts will be when the output is zero, we can find the π₯-intercepts by solving each factor is equal to zero.

And solving each of these factors to be equal to zero gives us π₯ is equal to zero or π₯ is equal to two or π₯ is equal to negative two. Weβre now ready to sketch the curve π¦ is equal to π prime of π₯. Weβll start by plotting the three π₯-intercepts onto our axes. And we see weβre sketching the cubic equation π¦ is equal to two π₯ cubed minus eight π₯. This polynomial has the leading term two π₯ cubed. Since the leading coefficient is two, this curve will have a similar shape to π¦ is equal to π₯ cubed. Using this, we get the following sketch of our curve π¦ is equal to π prime of π₯. Remember, we want to know when π prime of π₯ is less than zero and when π prime of π₯ is greater than zero.

From the sketch, we can see that π prime of π₯ is positive when π₯ is greater than two or when π₯ is between negative two and zero. So, we have π prime of π₯ is positive when π₯ is greater than negative two and less than zero or when π₯ is greater than two. And since the question wants us to write this in terms of intervals, this is the same as saying π is increasing on the open interval from negative two to zero and the open interval from two to β. We can do the same to check where our function is decreasing. We see from our sketch that our function π prime of π₯ is below the π₯-axis when π₯ is less than negative two or when π₯ is between zero and two.

So, weβve shown the derivative of π is less than zero when π₯ is less than negative two or when π₯ is greater than zero and less than two. And the question wants us to write this in terms of intervals. So, this is equivalent to saying that π is decreasing on the open interval from β to negative two and the open interval from zero to two. Therefore, weβve shown for the function π of π₯ is equal to π₯ to the fourth power over two minus four π₯ squared plus two is decreasing on the open interval from negative β to negative two and on the open interval from zero to two. And this function is increasing on the open interval from negative two to zero and on the open interval from two to β.