Video: Finding the Monotonicity Intervals of a Rational Function

Determine the intervals on which the function 𝑦 = 7π‘₯/(π‘₯ βˆ’ 8) is increasing and on which it is decreasing.

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Video Transcript

Determine the intervals on which the function 𝑦 is equal to seven π‘₯ divided by π‘₯ minus eight is increasing and on which it is decreasing.

The question wants us to find the intervals on which our function is increasing and the intervals on which it is decreasing. We can see that our function 𝑦 is equal to seven π‘₯ divided by π‘₯ minus eight is a rational function. And in particular, that tells us that this function is both continuous and differentiable on its domain. And we know how to check whether a differentiable function is increasing or decreasing. We know for a differentiable function 𝑓, if the slope of our function 𝑓 is positive on an interval, then our function 𝑓 is increasing on that interval. However, if the slope of our function is negative on an interval, then we know that our function 𝑓 is decreasing on that interval. So we can use this to find the intervals on which our rational function is increasing or decreasing.

Let’s start by finding the domain of our function. Since our function is rational, its domain will be all real numbers except where our denominator is equal to zero. And we can see that our denominator will be equal to zero when π‘₯ is equal to eight. So the domain of our rational function is all real numbers except where π‘₯ is equal to eight. We now need to find an expression for the derivative of our function so we can determine the intervals on which it’s increasing and decreasing. So we need to differentiate 𝑦 is equal to seven π‘₯ divided by π‘₯ minus eight with respect to π‘₯. There’s a few different ways of doing this. For example, we could divide seven π‘₯ by π‘₯ minus eight by using algebraic division. Then we just need to differentiate each term separately.

And this would work. However, we’re going to do this by using the quotient rule for differentiation which tells us the derivative of the quotient to two functions 𝑒 divided by 𝑣 is equal to 𝑣𝑒 prime minus 𝑒𝑣 prime divided by 𝑣 squared. So we’ll set 𝑒 to be the function in our numerator, that’s seven π‘₯, and 𝑣 to be the function in our denominator, that’s π‘₯ minus eight. Then by the quotient rule, we have d𝑦 by dπ‘₯ is equal to 𝑣𝑒 prime minus 𝑒𝑣 prime divided by 𝑣 squared. So to use this, we just need to find expressions for 𝑒 prime and 𝑣 prime.

Let’s start with 𝑒 prime. It’s the derivative of seven π‘₯ with respect to π‘₯. This is just a linear term, so its derivative is just the coefficient of π‘₯, which is seven. And we can do the same to find 𝑣 prime of π‘₯. It’s the derivative of π‘₯ minus eight with respect to π‘₯. But this is just a linear expression, so its derivative is the coefficient of π‘₯, which is just one. Substituting in our expressions for 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime, we get d𝑦 by dπ‘₯ is equal to π‘₯ minus eight times seven minus seven π‘₯ times one all divided by π‘₯ minus eight squared.

Distributing seven over our parentheses in our numerator, we get seven π‘₯ minus 56. Then we need to subtract seven π‘₯ times one, which is seven π‘₯. And then we can see that we can cancel seven π‘₯ minus seven π‘₯ to give us zero. So we’ve simplified this expression to show d𝑦 by dπ‘₯ is equal to negative 56 divided by π‘₯ minus eight squared. Now, all we need to do is look for intervals on which this expression is positive so that our function is increasing. And we need to find intervals on which this expression is negative to find where our function is decreasing.

Let’s start by looking at the denominator in our expression. We can see it’s π‘₯ minus eight all squared. And since π‘₯ is not allowed to be equal to eight, we’re always squaring a nonzero number. And the square of any nonzero number is greater than zero. So our denominator is always positive. Let’s now check our numerator. We’ll include the negative signs, so our numerator will be negative 56. And this is just a constant, so our numerator is negative.

So what we’ve shown is, for any value of π‘₯ in the domain of our function, the derivative of our function is a negative number divided by a positive number. And a negative number divided by a positive number is negative. So our slope is negative for all values of π‘₯ in our domain. So our function must be decreasing for all values of π‘₯ not equal to eight.

In other words, we’ve shown that our function is decreasing on the entire set of real numbers except for the point eight. And this is the entire domain of our function. So we don’t need to check what happens when π‘₯ is equal to eight. Therefore, given the function 𝑦 is equal to seven π‘₯ divided by π‘₯ minus eight, we’ve shown that this function is decreasing on its entire domain which is all real values of π‘₯ except when π‘₯ is equal to eight.

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