Video: Discussing the Monotonicity of a Rational Function

Which of the following statements is true for the function β„Ž(π‘₯) = (βˆ’1/(7 βˆ’ π‘₯)) βˆ’ 5? [A] β„Ž(π‘₯) is decreasing on the intervals (βˆ’βˆž, 7) and (7, ∞). [B] β„Ž(π‘₯) is decreasing on the intervals (βˆ’βˆž, βˆ’7) and (βˆ’7, ∞). [C] β„Ž(π‘₯) is increasing on the intervals (βˆ’βˆž, βˆ’7) and (βˆ’7, ∞). [D] β„Ž(π‘₯) is increasing on the intervals (βˆ’βˆž, 7) and (7, ∞).

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Video Transcript

Which of the following statements is true for the function β„Ž of π‘₯ equals negative one over seven minus π‘₯ minus five? Is it (A) β„Ž of π‘₯ is decreasing on the intervals negative ∞ to seven and seven to ∞? Is it (B) β„Ž of π‘₯ is decreasing on the intervals negative ∞ to negative seven and negative seven to ∞? (C) β„Ž of π‘₯ is increasing on the intervals negative ∞ to negative seven and negative seven to ∞. Or (D) β„Ž of π‘₯ is increasing on the intervals negative ∞ to seven and seven to ∞.

If we look carefully, we see that β„Ž of π‘₯ is a reciprocal function. It’s one over some polynomial. And so we know that there are probably going to be asymptotes on our graph. Let’s think about how we might sketch the graph of β„Ž of π‘₯. We’ll begin by starting with the function 𝑓 of π‘₯ is equal to one over π‘₯. And then we’re going to consider the series of transformations that map the function one over π‘₯ onto the function β„Ž of π‘₯. Here is the function one over π‘₯. It has horizontal and vertical asymptotes made up of the π‘₯- and 𝑦-axis. Now we’ll consider how we map 𝑓 of π‘₯ onto one over negative π‘₯. This is represented by reflection in the 𝑦-axis.

And then how do we map this onto the function one over seven minus π‘₯? Well, adding seven to the inner part of our composite function gives us a horizontal translation by negative seven. That’s a translation to the left seven units. Now, in doing this, our horizontal asymptote stays the same; it’s still the π‘₯-axis. But our vertical asymptote also shifts left seven units. And so it goes from being the 𝑦-axis, which is the line π‘₯ equals zero, to being the line π‘₯ equals negative seven. But of course, β„Ž of π‘₯ is negative one over seven minus π‘₯. This time, we reflect the graph in the 𝑦-axis. And so our horizontal asymptote remains unchanged, but our vertical asymptote is now at π‘₯ equals seven.

Our final transformation maps this function onto β„Ž of π‘₯. That’s negative one over seven minus π‘₯ minus five. And now we move the entire graph we translated five units down. And so we now have the graph of β„Ž of π‘₯ of negative one over seven minus π‘₯ minus five, and we’re ready to decide whether the function is increasing or decreasing over the various intervals. Remember, if a function is decreasing, its graph will have a negative slope, and if it’s increasing, its graph will have a positive slope. As we move our values of π‘₯ from left to right, that is, from negative ∞ all the way up to π‘₯ equals seven, we see that the graph is sloping downwards. It will approach negative ∞, but never quite reach it.

Then as π‘₯ approaches positive ∞ from seven, the graph continues to slope downwards. This time, though, it approaches negative five. And so the function is definitely decreasing over these intervals from negative ∞ to seven and seven to ∞. Since the function itself cannot take a value of π‘₯ equals seven, and this is why we have the horizontal asymptote, then we want to include open intervals. Those are the round brackets. And so the correct answer must be (A), β„Ž of π‘₯ is decreasing on the open interval negative ∞ to seven and seven to ∞.

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