Question Video: Finding the Intervals of Increasing and Decreasing of a Function Involving Logarithmic Functions Using the Chain Rule Mathematics • Higher Education

Find the intervals on which the function 𝑓(π‘₯) = 5 ln (βˆ’4 ln π‘₯ + 6) is increasing and decreasing.

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Video Transcript

Find the intervals on which the function 𝑓 of π‘₯ is equal to five times the natural logarithm of negative four times the natural logarithm of π‘₯ plus six is increasing and decreasing.

We’re given a function 𝑓 of π‘₯. And we need to determine the intervals on which this function is increasing and the intervals on which this function is decreasing. And the first thing we should always do when we’re asked a question like this is to determine the domain of our function 𝑓 of π‘₯. The reason we want to find the domain is we can’t possibly say that our function 𝑓 of π‘₯ is increasing or decreasing on an interval if our function 𝑓 of π‘₯ is not defined across the entire interval. So let’s determine the domain of our function 𝑓 of π‘₯.

First, we can see that our function 𝑓 of π‘₯ contains the natural logarithm of π‘₯. And the natural logarithm of π‘₯ is only defined when π‘₯ is positive. So we know that π‘₯ must be greater than zero. Next, we can also see we’re taking the natural logarithm of negative four times the natural logarithm of π‘₯ plus six. And if we’re taking the natural logarithm of this, then it must be positive. Otherwise, 𝑓 of π‘₯ will not be defined. And for any other value of π‘₯, we’re just taking the natural logarithm of positive numbers. So this will be defined. So we need to find the values of π‘₯ where negative four times the natural logarithm of π‘₯ plus six is positive.

We can just solve this inequality. We’ll start by subtracting six from both sides. We get negative four times the natural logarithm of π‘₯ is greater than negative six. Next, we’ll divide our inequality through by negative four. Remember, because this is a negative number, we need to switch the direction of our inequality. This gives us the natural logarithm of π‘₯ should be less than negative six divided by negative four, which we simplify to give us three over two. To solve this inequality, we need to recall that the exponential function is an increasing function. In other words, if the natural logarithm of π‘₯ is less than three over two, then 𝑒 to the power of the natural logarithm of π‘₯ will be less than 𝑒 to the power of three over two. Finally, we know the exponential function and the natural logarithm functions are inverses. So this gives us that π‘₯ will be less than 𝑒 to the power of three over two.

So we found our domain 𝑓 of π‘₯. π‘₯ must be positive; however, π‘₯ must be less than three over two. And because in the question, we’re working with intervals, we’ll write this in interval notation. The domain of 𝑓 of π‘₯ is the open interval from zero to 𝑒 to the power of three over two. Now that we found the domain of 𝑓 of π‘₯, we’re ready to start finding the intervals on which 𝑓 of π‘₯ is increasing and decreasing.

Let’s start by clearing some space. To answer this question, we need to recall the following piece of information about increasing and decreasing functions. We know if 𝑓 prime of π‘₯ is greater than zero on an interval 𝐼, then 𝑓 of π‘₯ will be increasing on that interval 𝐼. However, if 𝑓 prime of π‘₯ is less than zero on our interval 𝐼, then we know that 𝑓 of π‘₯ will be decreasing on that interval 𝐼. In other words, to determine where our function 𝑓 of π‘₯ is increasing or decreasing, we can instead look at the sign of 𝑓 prime of π‘₯. And to do this, we’re going to need to differentiate our function 𝑓 of π‘₯. We can see that 𝑓 of π‘₯ is the composition of two functions. So we’re going to want to do this by using the chain rule.

So let’s start by recalling the following version of the chain rule. This tells us if 𝑒 and 𝑣 are differentiable functions, then the derivative of 𝑒 of 𝑣 of π‘₯ with respect to π‘₯ is equal to 𝑣 prime of π‘₯ times 𝑒 prime of π‘₯ evaluated at 𝑣 of π‘₯. To apply this, we’re first going to need to write 𝑓 of π‘₯ as the composition of two functions. We’ll need to set 𝑣 of π‘₯ to be our inner function. We’ll set 𝑣 of π‘₯ to be negative four times the natural logarithm of π‘₯ plus six. So, by setting 𝑓 of π‘₯ equal to this, we can see that 𝑓 of π‘₯ is five times the natural logarithm of 𝑣. Then all we need to do is set our function 𝑒 of 𝑣 to be the remaining function. 𝑒 of 𝑣 is five times the natural logarithm of 𝑣.

And now we can see we’ve successfully written 𝑓 of π‘₯ as 𝑒 evaluated at 𝑣 of π‘₯. So we can find 𝑓 prime of π‘₯ using the chain rule. To use the chain rule, we’re going to need to find the expressions for 𝑣 prime of π‘₯ and 𝑒 prime of 𝑣. Let’s start with 𝑣 prime of π‘₯. That’s the derivative of negative four times the natural logarithm of π‘₯ plus six. We can differentiate this term by term. Remember, the derivative of the natural logarithm function is the reciprocal function. We get 𝑣 prime of π‘₯ is negative four divided by π‘₯. And in fact we can do the same to find 𝑒 prime of 𝑣; we get 𝑒 prime of 𝑣 is five divided by 𝑣.

We can now find an expression for 𝑓 prime of π‘₯ by substituting our expressions for 𝑣 prime and 𝑒 prime into the chain rule. We get 𝑓 prime of π‘₯ is equal to negative four over π‘₯ multiplied by five divided by 𝑣 of π‘₯. And of course, we know that 𝑣 of π‘₯ is negative four times the natural logarithm of π‘₯ plus six. So we substitute this in. This gives us the following expression. Finally, we’ll multiply these two expressions together to give us that 𝑓 prime of π‘₯ is equal to negative 20 divided by π‘₯ times negative four multiplied by the natural logarithm of π‘₯ plus six.

We now need to determine on which intervals is this positive and on which intervals is this negative. And remember, we’ve shown the domain of our function 𝑓 of π‘₯ is the open interval from zero to 𝑒 to the power of three over two and in particular on this interval π‘₯ is positive. So let’s look at our expression for 𝑓 prime of π‘₯. We have π‘₯, and we know that this is positive. The next question we need to ask is, what happens to negative four times the natural logarithm of π‘₯ plus six on this interval? There’s a few different ways of answering this question. For example, we could use a graphical approach; however, we’re going to use the fact that the natural logarithm of π‘₯ is an increasing function.

If the natural logarithm of π‘₯ is an increasing function, then on the open interval from zero to 𝑒 to the power of three over two, we must have the natural logarithm of π‘₯ is less than the natural logarithm of 𝑒 to the power of three over two. And this is because the natural logarithm is an increasing function. The higher our input of π‘₯, the bigger the output. So we just inputted the largest value in our interval. But we can evaluate the natural logarithm of 𝑒 to the power of three over two because the natural logarithm and exponential functions are inverse functions. It’s just equal to three over two. What we’ve shown is on the domain of 𝑓 of π‘₯, the natural logarithm of π‘₯ is less than three over two. And if the natural logarithm of π‘₯ is less than three over two, what does this mean about negative four times the natural logarithm of π‘₯ plus six?

And there’s a few different ways of answering this question. We’re going to do this by recreating this from our inequality. We’ll start by multiplying both sides of our inequality through by negative four. And of course, because this is negative, this switches the direction of our inequality. We have negative four times the natural logarithm of π‘₯ must be bigger than negative six. Now all we need to do is add six to both sides of our inequality. This gives us that negative four times the natural logarithm of π‘₯ plus six is positive on the domain of 𝑓 of π‘₯. So on the entire domain of 𝑓 of π‘₯, negative four times the natural logarithm of π‘₯ plus six is positive. So, in fact, what we’ve shown is on the entire domain of 𝑓 of π‘₯, 𝑓 prime of π‘₯ is a negative number divided by the product of two positive numbers.

Therefore, 𝑓 prime of π‘₯ is negative on the entire domain of 𝑓 of π‘₯. And we know if 𝑓 prime of π‘₯ is negative on the entire domain of 𝑓 of π‘₯, then 𝑓 of π‘₯ is decreasing on the entire domain of 𝑓 of π‘₯. And of course, this is setting 𝐼 equal to the domain of 𝑓 π‘₯, which we’ve shown is the open interval from zero to 𝑒 to the power of three over two.

Therefore, we were able to show for the function 𝑓 of π‘₯ is equal to five times the natural logarithm of negative four multiplied by the natural logarithm of π‘₯ plus six is decreasing on the open interval from zero to 𝑒 to the power of three over two.

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