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

Let 𝑓(π‘₯) = 3π‘₯⁴𝑒^(βˆ’4π‘₯). Determine the intervals where this function is increasing and where it is decreasing.

04:38

Video Transcript

Let 𝑓 of π‘₯ be equal to three π‘₯ to the fourth power multiplied by 𝑒 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 𝑓 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 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 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, 𝑓 of π‘₯ is a differentiable function.

So, we’ll start by finding our derivative function 𝑓 prime of π‘₯. Since 𝑓 of π‘₯ 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 𝑒 and 𝑣 with respect to π‘₯ is equal to 𝑒 prime 𝑣 plus 𝑒𝑣 prime. So, to differentiate 𝑓 of π‘₯, we’ll set 𝑒 equal to three π‘₯ to the fourth power and 𝑣 equal to 𝑒 to the power of negative four π‘₯.

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

We can simplify three π‘₯ to the fourth power multiplied by negative four 𝑒 to the power of negative four π‘₯ as negative 12π‘₯ to the fourth power times 𝑒 to the power of negative four π‘₯. So, we now have an expression for our derivative function 𝑓 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 𝑒 to the power of negative four π‘₯. So, if we take this factor out, we see that our derivative function 𝑓 prime of π‘₯ is equal to 12π‘₯ cubed times 𝑒 to the power of negative four π‘₯ multiplied by one minus π‘₯.

And remember, we’re looking for the values of π‘₯ where 𝑓 prime of π‘₯ is greater than zero and the values of π‘₯ where 𝑓 prime of π‘₯ 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 12π‘₯ cubed times 𝑒 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‘₯, 𝑒 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 it’s 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 𝑦 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 𝑓 prime of π‘₯ is positive or negative. So, we’ve shown that 𝑓 prime of π‘₯ is greater than zero when π‘₯ is greater than zero or π‘₯ is less than one and 𝑓 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 ∞ to zero. And saying that π‘₯ is greater than one is the same as saying π‘₯ is in the open interval from one to ∞. Therefore, we’ve shown if 𝑓 of π‘₯ is equal to three π‘₯ to the fourth power times 𝑒 to the power of negative four π‘₯. 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 ∞.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.