Answer: $C = 6$

In order for the function to be continuous, we must have $\displaystyle{\lim_{x \to 3}f(x) = f(3) = C}$. So let’s find that limit:

\begin{align*} \lim_{x \to 3} \frac{x^2 – 9}{x-3} &= \lim_{x \to 3}\frac{(x+3)(x-3)}{(x-3)} \\ &= \lim_{x \to 3}\frac{(x+3)\cancel{(x-3)}}{\cancel{(x-3)}}\\ &= \lim_{x \to 3}x+3 = 6 \end{align*} Hence $\displaystyle{C = \lim_{x \to 3}f(x) = 6} \quad \cmark$

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Many students have no idea what to do when faced with this brief question. Here’s how to approach it.

First, note that the question is not asking you to find this number; instead, it is asking whether such a number exists. And that is a BIG CLUE we should think about the Intermediate Value Theorem (IVT). With that in mind, we start by expressing the statement mathematically:

Is there some number, $x$, that is exactly 1 more than its cube, $x^3$. That is, is there a solution to the equation \[x = x^3 + 1 \text{ ?}\] As we noted above, when asked whether two expressions are ever equal, it’s helpful to define a new function that is the difference between those two expressions. Hence we define $f$: \begin{align*} f(x) &= x^3 + 1 – x \\ \\ &= x^3 -x +1 \end{align*} The question now becomes whether there exists a solution to $f(x) = 0$.

Note that since $f(x)$ is a polynomial, it is continuous. Hence we can use the IVT to prove that there exists a solution to $f(x) = 0$. . . if we can find some number $a$ such that $f(a) \gt 0$, and some number $b$ such that $f(b) \lt 0.$ Then, by the IVT we know that there is a value $c$ between $a$ and $b$ for which $f(c) = 0$, and we’re done!

So now, we just try out some numbers and see what happens. The easiest number to try is 0: \[f(0) = 1 \text{, which is } > 0\] So we already have a value for $a$ such that $f(a) > 0$.

Now let’s look for $b$ such that $f(b) \lt 0$. The next-easiest number to try is 1: \[f(1) = 1 -1 + 1 = 1 \text{, which is also } > 0\] and so doesn’t work. Let’s next try $-1$: \[f(-1) = -1+1 + 1 = 1 \text{, which is again } > 0\] and so again doesn’t advance us.

Having exhausted those really-easy numbers to try, let’s stop and think a bit: we want $f(x)$ to be negative, and we know we substitute larger values, $x^3$ will dominate over $x$. Hence we’ll try only negative numbers from now on. Let’s try the next-easiest negative number, $-2$: \[f(-2) = -8 + 2 + 1 = -5 \text{ which is } \lt 0\] So that works as $b$ such that $f(b) \lt 0$! We’re essentially done, but should wrap things up by summarizing:

“For the function $f(x)$ defined above, $f(0) > 0$ and $f(-2) \lt 0$, and so by the IVT there exists a number $c$, $-2 \lt c \lt 0$, such that $f(c)=0$. Hence there exists a number $c$ that is exactly 1 more than its cube.” $\quad \cmark$

Furthermore, although the question doesn’t ask us to go any further, we actually have shown that the value of x we’re after lies in the range $[-2, -1].$

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When considering a function like this, that depends on whether $x$ is rational or irrational, the key thing to remember is that there are both rational and irrational numbers arbitrarily close to any given number $c$. No matter how close you get, there are always both rational and irrational numbers that lie between you and $c$.

Hence if you were to walk toward $x=c$ from either direction, you would just keep leaping between 0 and 1:

$$\displaystyle{\lim_{x \to c}f(x)} \text{ does not exist}$$ Thus $f(x)$ is not continuous for any values of x. $\quad \cmark$

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Before we use the theorem to prove the statement, let’s imagine a slightly different situation to develop an intuitive understanding of what’s going on.

Picture a hiker, Fran, who starts at the bottom of the mountain at 6:00 a.m and hikes up along the only existing path toward the top. Also imagine a second hiker, Greg, who starts his hike simultaneously with Fran at 6:00 a.m.—but Greg starts his hike from the top of the mountain, and heads down along the same single path. Fran finishes her hike at 6:00 p.m., arriving at the top of the mountain; Greg also finishes his hike at 6:00 p.m., arriving at the bottom of the mountain at that same moment. Now clearly Fran and Greg must pass each other at some point during the hike, since they’re on the same path and both hike for exactly the same 12 hours. We don’t know when their paths cross; indeed, the exact moment depends on how fast each hikes, when they take their breaks, and other details of their separate walks. But regardless of those details, we know that there is some moment when they cross paths.

The argument remains the same if we replace Greg’s hike with Fran’s return trip the following day: since she starts at 6:00 a.m. and finishes at 6:00 p.m., and follows the same path, there must be some moment when she is at the same location she was the day before at that time. (If it’s easier to picture, imagine the downward-headed Fran meeting her upward-headed twin who left the bottom of the mountain at 6:00 a.m. Their paths must cross at some point, even if we don’t know exactly when.)

To make our argument more formal, let the function $f(t)$ represent Fran’s distance along the path, as measured from the bottom of the mountain, on the first day or her hike. Hence $f$(6:00 a.m.) = 0, since she starts at the bottom of the mountain. Let’s say that when she reaches the top of the mountain, she has traveled a distance $D$ along the path, such that $f$(6:00 p.m.) = $D$. (See the graph: one possibility for $f(t)$ is shown in brown.) Note that $f(t)$ is a continuous function, since Fran moves continuously along the path: she can’t suddenly disappear from one spot and suddenly appear in another without traversing the points in-between.

Now let the function $g(t)$ represent Fran’s distance along the path, as measured from the bottom of the mountain, on the second day of her hike, when she takes her return trip. (We first imagined this above as Greg’s hike down the mountain.) Since she starts this hike at 6:00 a.m. at the top of the mountain, $g$(6:00 a.m.) = $D$. At the end of her hike, we have $g$(6:00 p.m.) = 0. (One possibility for this function is shown on the graph in blue.) Note that $g(t)$ is also a continuous function, for the same reason $f(t)$ is.

As you can see in the graphic, no matter the exact shape of two curves representing the two functions, they must cross at some point. We can’t say exactly where they cross without more information, but we know they must cross somewhere.

To use the Intermediate Value Theorem, let’s invoke an approach we’ve now used several times above, and create a new function, $h(t)$, that is the difference of the two functions above:

$$h(t) = g(t) – f(t)$$ Now, at the start of the day, $h(\text{6:00 a.m.}) = g(\text{6:00 a.m.}) – f(\text{6:00 a.m.})= D – 0 = D.$
By contrast, at the end of the day $h(\text{6:00 p.m.}) = g(\text{6:00 p.m.}) – f(\text{6:00 p.m.})= 0 – D = -D.$
Since $0$ is between $D$ and $-D$, by the Intermediate Value Theorem there must be a time $T$ between 6:00 a.m. and 6:00 p.m. such that $h(T) = 0$. That is, there must be some instant when the two functions have the same value—which means that she is at the same spot on the path at exactly the same time of the day on both days. Again, we would need more information to know when that time of day is, but regardless, we know that it must exist. $\quad \cmark$

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We don’t find the result that the question is asking to prove to be intuitive at all; in fact we find it remarkable. It is nonetheless true.

Before we follow the hint and launch into the proof, we can develop some understanding by considering two points on the opposite sides of equator, using the function that the hint defines:

$$f(\theta) = T(\theta + \pi) – T(\theta)$$ For simplicity, let’s first consider the point with longitude $\theta_a = 0$ radians (in the Atlantic Ocean); then its antipodal point has longitude $\theta_a + \pi = \pi$ radians (in the Pacific Ocean). Furthermore, let’s suppose for the moment that $T(0) > T(\pi)$. With that assumption, the difference between the two functions is negative: $$f(0) = T(\pi) – T(0) < 0$$ (We may not actually have $T(0) > T(\pi)$; it won’t matter for the argument, so let’s just suppose that it is.)

Now imagine increasing $\theta$, bit by bit, sweeping around the Earth’s circumference. Since the temperature varies continuously as we move around the circle, $f(\theta)$ varies continuously too.

We keep sweeping around until we come to the opposite side of the Earth, where $\theta_b = \pi$. But note that we’re comparing the same two points as we did before—the points with longitude $\pi$ radians (in the Pacific) and $2\pi$ radians (which is the same as 0 radians, back where we started in the Atlantic). But their order has reversed when we compute $f(\theta)$:

\begin{align*} f(\pi) &= T(2\pi) – T(\pi) &&\text{[Note that $T(2\pi) = T(0)$]}\\ \\ &= T(0) – T(\pi) \\ \\ &= -f(0) \\ \\ &> 0 \end{align*} So we started with $f < 0$, and swept around to where $f > 0$, and hence (since $f$ is continuous) by the Intermediate Value Theorem there must be a point somewhere between 0 and $\pi$ radians along the equator such that $f = 0$, i.e., where that point and its antipodal point have the same temperature.

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If you were to encounter this or a similar question on an exam, we suggest just jumping in with the hint and seeing where it takes you. (Don’t just stare at the blank page; start by writing down something that the problem told you!) The hint was $$f(\theta) = T(\theta + \pi) – T(\theta)$$ Since temperature is a continuous function on the Earth’s surface, $f(\theta)$ is a continuous function.

With no further guidance provided by the hint, let’s write an expression for $f$ at some particular angle alpha: $\theta = \alpha$. $$f(\alpha) = T(\alpha + \pi) – T(\alpha) \; \blacktriangleleft $$ Since that expression contains $T(\alpha + \pi)$, let’s also write the expression for $f(\alpha + \pi)$ just to see what happens: $$f(\alpha + \pi) = T(\alpha + 2\pi) – T(\alpha + \pi)$$ But note that $T(\alpha + 2\pi) = T(\alpha)$ since $T$ is a periodic function with period $2\pi$. (That is, we’ve just swept all the way around the circle to back where we started.) Hence we can rewrite the preceding line:

$$f(\alpha + \pi) = T(\alpha) – T(\alpha + \pi)$$ Comparing that to our expression for $f(\alpha)$ above $(\blacktriangleleft)$, we see that $$f(\alpha + \pi) = -f(\alpha)$$ Let’s consider the possibilities here: If both $f(\alpha + \pi)$ and $f(\alpha)$ are zero, then that particular value of $\alpha$ specifies a point that has the same temperature as its antipodal point at $\alpha + \pi$, and we’ve shown what we were after.

If they are not zero and $f(\alpha)$ is positive, then the last expression tells us that $f(\alpha + \pi)$ is negative; if $f(\alpha)$ is negative, then $f(\alpha + \pi)$ is positive. Hence by the Intermediate Value Theorem, there must exist an angle we’ll call $\theta_c$ in $(0, \pi)$ for which $f(\theta_c) = 0$, and hence where the point specified by $\theta_c$ and its antipodal point at $\theta_c + \pi$ have the same temperature. $\quad \cmark$

As an aside, this remarkable fact about two anitpodal points having the same temperature is a specific application of something called the Borsuk–Ulam Theorem.

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