Question Video: Sketching the Graph of Rational Functions Using Derivatives Mathematics • Higher Education

Sketch the graph of 𝑓(𝑥) = 3𝑥²/(𝑥² − 4).

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Video Transcript

Sketch the graph of 𝑓 of 𝑥 equals three 𝑥 squared over 𝑥 squared minus four.

Let’s begin by considering the domain and range of our function. We know that a function which is a quotient won’t exist at points where the denominator of the quotient is zero. So we set 𝑥 squared minus four equals zero and solve for 𝑥 to find the domain of our function. We can add four to both sides of this equation. Then we take the square root of both sides of our equation, remembering to take both the positive and negative square root of four. And we obtain that when 𝑥 squared minus four is equal to zero, 𝑥 is equal to positive or negative two. So the domain of our function is all real values, except 𝑥 is equal to positive or negative two. Now the range itself is a set of possible outputs on the function. And we can infer this through the graphing process.

Next, we’ll work out whether there are any intercepts. By setting 𝑥 equal to zero and solving for 𝑦, we’ll find the 𝑦-intercept. When we do, we obtain 𝑦 to be equal to three times zero squared over zero squared minus four, which is zero. So there’s a 𝑦-intercept at 𝑦 equals zero. Next, we set 𝑦 or 𝑓 of 𝑥 equal to zero and solve for 𝑥. That is, zero equals three 𝑥 squared over 𝑥 squared minus four.

Now for this to be true, we know that the numerator of this fraction must itself be equal to zero. And for three 𝑥 squared to be equal to zero, 𝑥 must be equal to zero. So we actually only have one intercept altogether. And that’s at the origin: zero, zero. Next, we can check for symmetry. An even function is one for which 𝑓 of negative 𝑥 equals 𝑓 of 𝑥, whereas an odd function is one for which 𝑓 of negative 𝑥 equals negative 𝑓 of 𝑥. Well, 𝑓 of negative 𝑥 is three times negative 𝑥 squared over negative 𝑥 squared minus four, which is equal to 𝑓 of 𝑥. So our function is even. And that means it’s going to be symmetrical about the 𝑦-axis.

Next, we’ll look for asymptotes. We’ll look for horizontal asymptotes by considering what happens to our function as 𝑥 approaches either positive or negative infinity. We can’t use direct substitution as when we substitute in 𝑥 equals either positive or negative infinity, we end up with infinity over infinity, which is undefined. So instead, we divide both the numerator and the denominator of our expression by 𝑥 squared. That gives us three over one minus four over 𝑥 squared.

And now we can use direct substitution. As 𝑥 approaches either positive or negative infinity, four over 𝑥 squared approaches zero. So the limit of our function is three over one, which is just three. And we see that the line 𝑦 equals three must be a horizontal asymptote.

Remember, we said that when 𝑥 is equal to positive or negative two, the denominator is zero. So we find the following limits. The limit as 𝑥 approaches two from the right of our function is infinity. And the limit as 𝑥 approaches two from the left of the function is negative infinity. And the limit as 𝑥 approaches negative two from the right of the function is negative infinity. And as it approaches negative two from the left, it’s positive infinity. So we obtain 𝑥 equals positive or negative two to be vertical asymptotes.

Let’s clear some space and look for intervals of increase and decrease. We’ll begin by using the quotient rule to find the first derivative of our function. By letting 𝑢 be equal to three 𝑥 squared and 𝑣 be equal to 𝑥 squared minus four, we obtain expressions for d𝑢 by d𝑥 and d𝑣 by d𝑥. And so the first derivative of our function is 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 over 𝑣 squared. And distributing our parentheses, we find the first derivative to be negative 24𝑥 over 𝑥 squared minus four all squared.

Now since 𝑥 squared minus four all squared must be greater than zero for all 𝑥, this means the first derivative of our function must be less than zero when 𝑥 is greater than zero. So for values of 𝑥 greater than zero, we get negative 24 multiplied by a positive, which gives us a negative value. And so for 𝑥 greater than zero, the first derivative is less than zero. The opposite is true. When 𝑥 is less than zero, the first derivative is greater than zero. Since we know the function doesn’t exist when 𝑥 is equal to negative two, we can say that its intervals of increase are negative infinity to negative two and negative two to zero. And it has open intervals of decrease over zero to two and two to infinity.

Now, actually, if we look carefully, we also see that the first derivative is equal to zero when 𝑥 is equal to zero. So there’s a critical point at the point zero, zero. Coincidently, that’s also the point where our curves cross the axes.

We now have everything we need to complete our sketch. We know that there’s a horizontal asymptote at 𝑦 equals three and vertical asymptotes at 𝑥 equals negative two and 𝑥 equals two. There’s a critical point and an intercept at zero, zero. And that’s important because we see that the curve won’t be able to cross the 𝑥-axis here or here. That helps us decide where these two parts of the curve go. The curve is increasing over the open interval negative infinity to negative two and decreasing over the open interval two to infinity. And of course, we have these asymptotes. The function is also even and symmetrical about the 𝑦-axis. So this confirms we’re probably on the right track. We then use the remaining pieces of information to sketch the final part of our curve. And we’ve successfully sketched the graph of 𝑓 of 𝑥 equals three 𝑥 squared over 𝑥 squared minus four.

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