Video: Differentiating Rational Functions at a Point Using the Quotient Rule

If the function 𝑓(π‘₯) = (5π‘₯ + 7)/(4π‘₯ + 2), determine its rate of change when π‘₯ = 2.

02:09

Video Transcript

If the function 𝑓 of π‘₯ equals five π‘₯ plus seven over four π‘₯ plus two, determine its rate of change when π‘₯ is equal to two.

We recall the definition for the rate of change of the function or its derivative. It’s the limit as β„Ž approaches zero of 𝑓 of π‘Ž plus β„Ž minus 𝑓 of π‘Ž over β„Ž assuming that limit exists. In this question, 𝑓 of π‘₯ is equal to five π‘₯ plus seven over four π‘₯ plus two. And we want to find the rate of change when π‘₯ is equal to two. So we’re going to let π‘Ž be equal to two. So we need to evaluate 𝑓 prime of two, the rate of change or the derivative of our function, when π‘₯ equals two.

By our definition, it’s the limit as β„Ž approaches zero of 𝑓 of two plus β„Ž minus 𝑓 of two over β„Ž. Let’s work out 𝑓 of two plus β„Ž and 𝑓 of two. To find 𝑓 of two plus β„Ž, we replace each instance of π‘₯ in our original function with two plus β„Ž. And we get five times two plus β„Ž plus seven over four times two plus β„Ž plus two. And when we distribute our parentheses and simplify, we get 17 plus five β„Ž over 10 plus four β„Ž. Similarly, 𝑓 of two is five times two plus seven over four times two plus two which is 17 tenths. And we now see that 𝑓 prime of two is the limit as β„Ž approaches zero of the difference between these all over β„Ž.

There are two fractions in our numerator. So we’re going to simplify by creating a common denominator there. We’ll multiply the numerator and denominator of the first fraction on the numerator by 10 and the second fraction on our numerator by 10 plus four β„Ž. And when we do, we achieve the numerator shown. Well, this doesn’t make a lot of sense. But actually, dividing this entire fraction by β„Ž is the same as timesing it by one over β„Ž. So we rewrite our denominator as a β„Ž times 100 plus 40β„Ž. And then, we simplify a numerator to negative 18β„Ž. And you might now see that we can simplify further by dividing through by β„Ž.

And we’re now ready to perform direct substitution. By replacing β„Ž with zero, we find that 𝑓 prime of two is negative 18 over 100, which simplifies to negative nine over 50. The rate of change of our function 𝑓 of π‘₯ when π‘₯ is equal to two is negative nine over 50.

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