# Rational word problem: comparing two rational functions

So far we have learned how to solve rational equations, inequalities, and rates of change. I have not encountered a word problem like this and am unsure where to start. Graphing calculators are not allowed but I did look at the graphs for both functions and they both have the same asymptotes and origin at (0,0) and share the points (-1.4142, -5.657) and (1.4142, 5.657). I am not sure how to proceed and solve this algebraically.

Thank you very much for your time and help.

In a chemistry class, the students in lab derived a function to model the results of their experiment on the effect of heat on a chemical where $x$ represents the number of minutes heat was applied at a constant temperature set by the lab instructions. Their function was $f(x) = \frac{16x}{x^2 + 2}$. The teacher said the function should have been $f(x) = \frac{12x}{x^2 + 1}$.

a) Was there ever any time at which these two functions were the same. If so, when?

b) For what values of $x$ is their derived function greater than the actual function?

c) Estimate the instantaneous rate of change of each function at the time when they are equal.

• I realize now that it says "estimate" the instantaneous rate of change. This suggests that you are not quite trying to find the derivative. I would suggest finding the values of your two functions at $x = 1$ and $x = 2$ and then calculating the straight-line slope, $m = \frac{\Delta y}{\Delta x}$, and using that as your estimation (likewise for $x = 0$ you could use either $x = -1$ and $x = 1$ or even just $x = 0$ and $x = 1$. – Jared May 4 '18 at 3:46

When solving rational functions, you should (although not always necessary), setup the equation, set it equal to zero, find a common denominator and simplify the two fractions so that it's a single fraction, then solve the equation of the numerator set equal to zero:

$$\frac{16x}{x^2 + 2} - \frac{12x}{x^2 + 1} = \frac{16x(x^2 + 1) - 12x(x^2 + 1)}{(x^2 + 2)(x^2 + 1)} = 0$$

This is, initially, a cubic function (meaning, in general, it will be difficult to solve by hand). However, you can very quickly factor out an $x$ (meaning that $x = 0$ is a solution) and then it becomes a pretty simple quadratic:

$$x(16x^2 + 16 - 12x^2 - 24) = 0 \leadsto x(4x^2 - 8) = 0 \leadsto x^2 = 2$$

That makes sense with what you found since: $x = \pm\sqrt{2} \approx \pm 1.414$.

Now that you have the critical points ($x = -\sqrt{2},\ 0,\ \sqrt{2}$), create a sign chart to find when to find when the numerator is positive and negative. You can just choose numbers inside each of the four intervals (e.g. $x = -2,\ -1,\ 1,\ 2$). Then see whether or not your derived function is bigger or smaller for each of those points.

As for estimating the instantaneous rate of change. I'm not sure how you're supposed to do that, I would assume by using the definition of the limit (and not derivative rules from calculus).

The answer from c) will reinforce your answers from b) because the derivative will help make sense of why (or why not) there is a sign change. There are three cases:

1. $f_1'(x) < f_2'(x)$: The first function is changing at a slower (or less positive) rate, thus $f_2(x)$ will "overtake" $f_1(x)$. You would expect that "to the left", $f_1(x) > f_2(x)$ and "to the right", $f_1(x) < f_2(x)$.
2. $f_1'(x) > f_2'(x)$: This is the opposite of the first case.
3. $f_1'(x) = f_2'(x)$: They are changing at the same rate, so you would not expect a sign change in this case: if "to the left" $f_1(x) < f_2(x)$ then you'd expect it to be the same "to the right", or vice versa if "to the left" $f_1(x) > f_2(x)$ then so too would it be "to the right".
• Hi Jared, Thank you so much for your clear and detailed answer. I have followed through with your answer and it makes a lot more sense now. I am sorry if this is very obvious, but how did you cancel out the denominator (x^2+2)(x^2+1). I simplified until x(4x^2 - 8) / (x^2+2) (x^2+1). I am not sure how you simplified the equation after to get rid of the x out side of the bracket for the numerator and to cancel out the denominator. As for (c) where they ask for instantaneous ROC, my assumption is we use the points (1.4142) and (-1.4142) and input into the simplified equation we just calculated? – user554754 May 4 '18 at 2:10
• @user554754 Once you have a single fraction, you have $\frac{f(x)}{g(x)} = 0$. For this to be zero all you need is that $f(x)$, the numerator, is zero (since zero divided by anything is zero). Likewise, I "got rid" of the $x$ for the same reason. If you have $f(x)g(x)h(x) = 0$ then this will be zero when $f(x) = 0$ or $g(x) = 0$ or $h(x) = 0$, etc. Again, because anything multiplied by zero is zero. And no, you cannot just plug into your simplified equation to find the instantaneous rate of change--that will just give zero. – Jared May 4 '18 at 2:18
• @user554754 BTW, for the given situation, we should actually discard all negatives, since negatives don't fit with the domain of the word problem: "...$x$ represents the number of minutes heat was applied...". You can't apply heat for a negative amount of minutes. – Jared May 4 '18 at 2:23
• I am weak on theory, thank you for explaining clearly. I have a much better understanding of how to solve such questions. It was similar to finding points of intersection for two lines where both equations are set to equal one and another; I had to analyze the question better. Appreciate your efforts. – user554754 May 4 '18 at 12:49