Video: Finding the Intervals of Increasing and Decreasing of a Function Involving Using the Product Rule with Exponential Functions

Video Transcript

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

The question gives us the function 𝑓 of 𝑥 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 𝑓, we say that 𝑓 is increasing when its derivative is greater than zero. And it’s decreasing when its derivative is less than zero. And we’re given the function 𝑓 of 𝑥 which is the product of a polynomial function and an exponential function. Since polynomial functions and exponential functions are differentiable for all real values 𝑥, their product will also be differential for all real values 𝑥. So, 𝑓 of 𝑥 is a differentiable function.

So, we’ll start by finding our derivative function 𝑓 prime of 𝑥. Since 𝑓 of 𝑥 is the product of two functions, we’ll differentiate this by using the product rule. The product rule tells us, the derivative of the product of 𝑢 and 𝑣 with respect to 𝑥 is equal to 𝑢 prime 𝑣 plus 𝑢𝑣 prime. So, to differentiate 𝑓 of 𝑥, we’ll set 𝑢 equal to three 𝑥 to the fourth power and 𝑣 equal to 𝑒 to the power of negative four 𝑥.

To differentiate 𝑢, we use the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. This gives us 𝑢 prime is equal to 12𝑥 cubed. And for any constant 𝑛, we know the derivative of 𝑒 to the power of 𝑛𝑥 with respect to 𝑥 is equal to 𝑛 times 𝑒 to the power of 𝑛𝑥. This gives us that 𝑣 prime is equal to negative four times 𝑒 to the power of negative four 𝑥. So, by applying the product rule, we have that 𝑓 prime of 𝑥 is equal to 12𝑥 cubed times 𝑒 to the power of negative four 𝑥 plus three 𝑥 to the fourth power times negative four 𝑒 to the power of negative four 𝑥.

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

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

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

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

Finally, the question wants us to represent these as intervals. Saying that 𝑥 is greater than zero and 𝑥 is less than one is the same as saying 𝑥 is in the open interval from zero to one. Saying 𝑥 is less than zero is the same as saying 𝑥 is in the open interval from negative ∞ to zero. And saying that 𝑥 is greater than one is the same as saying 𝑥 is in the open interval from one to ∞. Therefore, we’ve shown if 𝑓 of 𝑥 is equal to three 𝑥 to the fourth power times 𝑒 to the power of negative four 𝑥. Then 𝑓 is increasing on the open interval from zero to one and decreasing on the open intervals from negative ∞ to zero and from one to ∞.