### Video Transcript

Determine the intervals on which the function ๐ of ๐ฅ equals eight ๐ฅ minus seven over seven ๐ฅ minus five is increasing and where it is decreasing.

Letโs begin by recalling how we identify whether a function is increasing or decreasing at a given point. We say that a function is increasing at a given point if the first derivative evaluated at that point is greater than zero. Similarly, the function will be decreasing at the point if the first derivative evaluated there is less than zero. Now, this question asks us to determine the intervals on which our function is increasing and decreasing, so essentially, all values of ๐ฅ such that ๐ prime of ๐ฅ is greater than zero or less than zero.

So, letโs begin then by evaluating the first derivative. Firstly, we notice that our function is the quotient of two functions. And so, weโre going to use the quotient rule to find its derivative. This says that the derivative of the quotient of two differentiable functions ๐ข and ๐ฃ is ๐ฃ times d๐ข by d๐ฅ minus ๐ข times d๐ฃ by d๐ฅ all over ๐ฃ squared.

Now, we know that our function is made up of the quotient of two differentiable functions because both the numerator and denominator are themselves polynomial functions. And we know polynomials are differentiable over their entire domain. And so, weโre going to define ๐ข as eight ๐ฅ minus seven, since thatโs the numerator of our fraction, and ๐ฃ as seven ๐ฅ minus five.

We see from our formula that weโre going to need to work out d๐ข by d๐ฅ and d๐ฃ by d๐ฅ. And so, weโll differentiate ๐ข and ๐ฃ term by term. Now, the derivative of eight ๐ฅ with respect to ๐ฅ is eight, and the derivative of a constant is zero. So, d๐ข by d๐ฅ is eight. Similarly, d๐ฃ by d๐ฅ is seven add zero, which is simply seven. And then, we see that the derivative of ๐ with respect ๐ฅ is ๐ฃ times d๐ข by d๐ฅ minus ๐ข times d๐ฃ by d๐ฅ all over ๐ฃ squared. Now, we are going to leave the denominator as it is. And weโre simply going to distribute the parentheses on our numerator and simplify.

We notice that seven ๐ฅ times eight minus eight ๐ฅ times seven gives us zero. And so, weโre going to work out negative five times eight minus negative seven times seven. Well, thatโs negative 40 minus negative 49, which is simply nine. And so, weโve found an expression for the first derivative of our function. Itโs nine over seven ๐ฅ minus five squared.

To find the intervals on which our function is increasing and decreasing, we want to find the values of ๐ฅ such that this is greater than zero and less than zero, respectively. Now, we can actually use a little bit of logic to do so. We begin by noticing that the numerator of our first derivative, nine, will always be greater than zero no matter the value of ๐ฅ.

Our denominator will always be greater than or equal to zero. And this is because, for real values of ๐ฅ, no matter the expression inside our parentheses, when we square it, weโll have a positive result. And so, assuming the denominator of our fraction, of our first derivative, is not zero, we know that the first derivative itself must be greater than zero.

So, letโs have a look at the denominator of our fraction. Which value of ๐ฅ will make it equal to zero? Well, seven ๐ฅ minus five squared will be equal to zero when seven ๐ฅ minus five is equal to zero. And if we solve for ๐ฅ by adding five to both sides and then dividing by seven, we find that ๐ฅ is equal to five-sevenths gives us a denominator of zero. And this means our function is increasing for all real values of ๐ฅ not including ๐ฅ equals five-sevenths.

Now, in fact, what does it actually mean if the denominator of our derivative is equal to zero here? Well, if the first derivative is equal to zero or it doesnโt exist, then we have a critical point. So, we know there is a critical point at ๐ฅ equals five-sevenths. And in fact, itโs an asymptote.