### Video Transcript

Let π of π₯ be equal to three π₯ to the fourth power multiplied by π to the power of negative four π₯. Determine the intervals where this function is increasing and where it is decreasing.

The question gives us the function π of π₯ which is the product of a polynomial and an exponential function. And it wants us to find the intervals where this function is increasing and decreasing. And we recall for a differentiable function π, we say that π is increasing when its derivative is greater than zero. And itβs decreasing when its derivative is less than zero. And weβre given the function π of π₯ which is the product of a polynomial function and an exponential function. Since polynomial functions and exponential functions are differentiable for all real values π₯, their product will also be differential for all real values π₯. So, π of π₯ is a differentiable function.

So, weβll start by finding our derivative function π prime of π₯. Since π of π₯ is the product of two functions, weβll differentiate this by using the product rule. The product rule tells us, the derivative of the product of π’ and π£ with respect to π₯ is equal to π’ prime π£ plus π’π£ prime. So, to differentiate π of π₯, weβll set π’ equal to three π₯ to the fourth power and π£ equal to π to the power of negative four π₯.

To differentiate π’, we use the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. This gives us π’ prime is equal to 12π₯ cubed. And for any constant π, we know the derivative of π to the power of ππ₯ with respect to π₯ is equal to π times π to the power of ππ₯. This gives us that π£ prime is equal to negative four times π to the power of negative four π₯. So, by applying the product rule, we have that π prime of π₯ is equal to 12π₯ cubed times π to the power of negative four π₯ plus three π₯ to the fourth power times negative four π to the power of negative four π₯.

We can simplify three π₯ to the fourth power multiplied by negative four π to the power of negative four π₯ as negative 12π₯ to the fourth power times π to the power of negative four π₯. So, we now have an expression for our derivative function π prime of π₯. However, we can simplify this by noticing that both terms share a factor of 12, a factor of π₯ cubed, and a factor of π to the power of negative four π₯. So, if we take this factor out, we see that our derivative function π prime of π₯ is equal to 12π₯ cubed times π to the power of negative four π₯ multiplied by one minus π₯.

And remember, weβre looking for the values of π₯ where π prime of π₯ is greater than zero and the values of π₯ where π prime of π₯ is less than zero. Letβs start by finding the interval where our function π is increasing. That means the derivative is greater than zero. So, we need to solve 12π₯ cubed times π to the power of negative four π₯ multiplied by one minus π₯ is greater than zero. We see that 12 is positive and for all values of π₯, π to the power of negative four π₯ is also positive. So, finding the values of π₯ where this product is positive is the same as finding the values of π₯ where π₯ cubed times one minus π₯ is positive.

And we see that this is a fully factored fourth-degree polynomial. So, we could sketch this. We can find the π₯-intercepts by solving the fact itβs equal to zero. This gives us π₯ equals zero and π₯ equals one. We can see for any negative input of π₯, we will get a negative output. And we can also see if π₯ is greater than one, we will also get a negative output. Finally, for values of π₯ between zero and one, we see that π₯ cubed and one minus π₯ are both positive, so we will get a positive output.

From the sketch, we can see that our curve π¦ is equal to π₯ cubed multiplied by one minus π₯ is below the π₯-axis when π₯ is less than zero or when π₯ is greater than one. And we can see that our curve is above the π₯-axis when π₯ is between zero and one. And we know that π₯ cubed multiplied by one minus π₯ being positive or negative is the same as saying whether our function π prime of π₯ is positive or negative. So, weβve shown that π prime of π₯ is greater than zero when π₯ is greater than zero or π₯ is less than one and π prime of π₯ is less than zero when π₯ is less than zero or π₯ is greater than one.

Finally, the question wants us to represent these as intervals. Saying that π₯ is greater than zero and π₯ is less than one is the same as saying π₯ is in the open interval from zero to one. Saying π₯ is less than zero is the same as saying π₯ is in the open interval from negative β to zero. And saying that π₯ is greater than one is the same as saying π₯ is in the open interval from one to β. Therefore, weβve shown if π of π₯ is equal to three π₯ to the fourth power times π to the power of negative four π₯. Then π is increasing on the open interval from zero to one and decreasing on the open intervals from negative β to zero and from one to β.