Rational word problem: comparing two rational functions

Question

This is a Grade 12 Advanced Functions (pre-calculus) word problem:

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.

d) How does your answer in c) reinforce your answer in b)?

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Answer 1

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".