Determine the intervals on which the function 𝑓 of 𝑥 equals eight 𝑥 minus seven over seven 𝑥 minus five is increasing and where it is decreasing.
Let’s 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 𝑥 such that 𝑓 prime of 𝑥 is greater than zero or less than zero.
So, let’s begin then by evaluating the first derivative. Firstly, we notice that our function is the quotient of two functions. And so, we’re going to use the quotient rule to find its derivative. This says that the derivative of the quotient of two differentiable functions 𝑢 and 𝑣 is 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 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, we’re going to define 𝑢 as eight 𝑥 minus seven, since that’s the numerator of our fraction, and 𝑣 as seven 𝑥 minus five.
We see from our formula that we’re going to need to work out d𝑢 by d𝑥 and d𝑣 by d𝑥. And so, we’ll differentiate 𝑢 and 𝑣 term by term. Now, the derivative of eight 𝑥 with respect to 𝑥 is eight, and the derivative of a constant is zero. So, d𝑢 by d𝑥 is eight. Similarly, d𝑣 by d𝑥 is seven add zero, which is simply seven. And then, we see that the derivative of 𝑓 with respect 𝑥 is 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 all over 𝑣 squared. Now, we are going to leave the denominator as it is. And we’re simply going to distribute the parentheses on our numerator and simplify.
We notice that seven 𝑥 times eight minus eight 𝑥 times seven gives us zero. And so, we’re going to work out negative five times eight minus negative seven times seven. Well, that’s negative 40 minus negative 49, which is simply nine. And so, we’ve found an expression for the first derivative of our function. It’s nine over seven 𝑥 minus five squared.
To find the intervals on which our function is increasing and decreasing, we want to find the values of 𝑥 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 𝑥.
Our denominator will always be greater than or equal to zero. And this is because, for real values of 𝑥, no matter the expression inside our parentheses, when we square it, we’ll 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, let’s have a look at the denominator of our fraction. Which value of 𝑥 will make it equal to zero? Well, seven 𝑥 minus five squared will be equal to zero when seven 𝑥 minus five is equal to zero. And if we solve for 𝑥 by adding five to both sides and then dividing by seven, we find that 𝑥 is equal to five-sevenths gives us a denominator of zero. And this means our function is increasing for all real values of 𝑥 not including 𝑥 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 doesn’t exist, then we have a critical point. So, we know there is a critical point at 𝑥 equals five-sevenths. And in fact, it’s an asymptote.