The Intermediate Value Theorem

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Here we see a consequence of a function being continuous.

The Intermediate Value Theorem should not be brushed off lightly. Once it is understood, it may seem “obvious,” but mathematicians should not underestimate its power.

Intermediate Value Theorem If $f$ is a continuous function for all $x$ in the closed interval $[a,b]$ and $d$ is between $f(a)$ and $f(b)$, then there is a number $c$ in $(a, b)$ such that $f(c) = d$.

If you are more of a visual person, you should imagine a continuous function, where you know the value of the function at two endpoints, $x=a$ and $x=b$, but you don’t really know what the function does between the points $x=a$ and $x=b$:

The Intermediate Value Theorem says that despite the fact that you don’t really know what the function is doing between the endpoints, a point $x=c$ exists and gives an intermediate value for $f$.

Now, let’s contrast this with a time when the conclusion of the Intermediate Value Theorem does not hold.

Consider the following situation,

and select all that are true:

$f$ is continuous on $(a,b)$. $f$ is continuous on $[a,b)$. $f$ is continuous on $(a,b]$. $f$ is continuous on $[a,b]$. There is a point $c$ in $[a,b]$ with $f(c) = d$.

Building on the question above, it is not difficult to see that each of the hypothesis of the Intermediate Value Theorem is necessary.

Let’s see the Intermediate Value Theorem in action.

Explain why the function $f(x)=x^3 + 3x^2+x-2$ has a zero between $0$ and $1$.

Since $f$ is a polynomial, we see that $f$ is continuous for all real numbers. We will apply the IVT here. Let $a=\answer [given]{0}$, $b=\answer [given]{1}$, and $d=\answer [given]{0}$, since $f(a)=f(0)=\answer [given]{-2}$, $f(b)=f(1)=\answer [given]{3}$, and $0$ is between $f(a)=-2$ and $f(b)=3$. Then by the Intermediate Value Theorem, there is a point $c$ in the interval $[0,1]$ such that $f(c)=d=\answer [given]{0}$.

This example also points the way to a simple method for approximating roots.

Approximate a root of $f(x) =x^3 + 3x^2+x-2$ between $0$ and $1$ to within one decimal place.

Again, since $f$ is a polynomial, we see that $f$ is continuous for all real numbers. Consider the table

\[ \begin{array}{c|c} x & f(x) \\ \hline 0.1 & -1.869\\ 0.2 &-1.672\\ 0.3 & -1.403\\ 0.4 & -1.056\\ 0.5 & -0.625\\ 0.6 & -0.104\\ 0.7 & 0.513 \end{array} \]

By the Intermediate Value Theorem, $f$ has a root between $0.6$ and $0.7$. Repeating the process

\[ \begin{array}{c|c} x & f(x) \\ \hline 0.61 &-0.046719\\ 0.62 & 0.011528 \end{array} \]

and using the Intermediate Value Theorem, we can conclude that $f$ has a root between $0.61$ and $0.62$, and the root is $0.6$ rounded to one decimal place.

The Intermediate Value Theorem can be used to show that curves cross:

Explain why the graphs of the functions $f(x) = x^2\ln (x)$ and $ g(x) = 2x\cos (\ln (x))$ intersect on the interval $[1,e]$.

To start, note that both $f$ and $g$ are continuous functions on the interval $[1,e]$, and hence $h = f-g$ is also a continuous function on the interval $[1,e]$. Now \begin{align*} h(1) &= f(1) - g(1) \\ &= (\answer [given]{1})^2\cdot \ln (\answer [given]{1}) - 2\cdot \answer [given]{1}\cdot \cos (\ln (\answer [given]{1}))\\ &= \answer [given]{-2}. \end{align*}

and in a similar fashion \begin{align*} h(e) &= f(e) - g(e) \\ &= e^2\cdot \answer [given]{1} - 2\cdot e\cos (\answer [given]{1}) \end{align*}

Since $e>2$ and $0 < \cos (1)<1$ we see that the expression above is positive. Therefore, $h(1)<0<h(e)$, and by the Intermediate Value Theorem, there exist a number $c$ in $(1,e)$ such that

\[ h(c)=0. \]

But this means that

\[ f(c)-g(c)=0, \]

and that $f(c)=g(c)$. Therefore, the curves $y=f(x)$ and $y=g(x)$ intersect at the point $(c,f(c))$.

We can see this point of intersection by looking at the graphs of $y=f(x)$ and $y=g(x)$ on the given interval.

\[ \graph [xmin=1,xmax=e,ymin=-10,ymax=10]{x^2\ln (x), 2x\cos (\ln (x))} \]

Now we move on to a more subtle example:

Suppose you have two cats, Roxy and Yuri. Is there a time when Roxy and Yuri have the same amount of water in their bowls assuming:

They start and finish drinking at the same times.
Roxy starts with more water than Yuri, and leaves less water left in her bowl than Yuri.

To solve this problem, consider two functions:

$W_{\mathrm {Roxy}}(t) =$ the amount of water in Roxy’s bowl at time $t$.
$W_{\mathrm {Yuri}}(t) =$ the amount of water in Yuri’s bowl at time $t$.

Now if $t_\mathrm {start}$ is the time the cats start drinking and $t_\mathrm {finish}$ is the time the cats finish drinking. Then we have

\[ W_\mathrm {Roxy}(t_\mathrm {start})-W_{\mathrm {Yuri}}(t_\mathrm {start}) > 0\]

and

\[ W_\mathrm {Roxy}(t_\mathrm {finish})-W_{\mathrm {Yuri}}(t_\mathrm {finish}) < 0. \]

Since the amount of water in a bowl at time $t$ is a continuous function, as water is “lapped” up in continuous amounts,

\[ W_\mathrm {Roxy}-W_{\mathrm {Yuri}} \]

is a continuous function, and hence the Intermediate Value Theorem applies. Since $W_\mathrm {Roxy}-W_{\mathrm {Yuri}}$ is positive when at $t_\mathrm {start}$ and negative at $t_\mathrm {finish}$, there is some time $t_\mathrm {equal}$ when the value is zero, meaning

\[ W_\mathrm {Roxy}(t_\mathrm {equal})- W_{\mathrm {Yuri}}(t_\mathrm {equal}) =0 \]

meaning there is the same amount of water in each of their bowls.

And finally, an example when the Intermediate Value Theorem does not apply.

Suppose you have two cats, Roxy and Yuri. Is there a time when Roxy and Yuri have the same amount of dry cat food in their bowls assuming:

They start and finish eating at the same times.
Roxy starts with more food than Yuri, and leaves less food uneaten than Yuri.

Here we could try the same approach as before, setting:

$F_{\mathrm {Roxy}}(t) =$ the amount of dry cat food in Roxy’s bowl at time $t$.
$F_{\mathrm {Yuri}}(t) =$ the amount of dry cat food in Yuri’s bowl at time $t$.

However in this case, the amount of food in a bowl at time $t$ is not a continuous function! This is because dry cat food consists of discrete kibbles, and is not eaten in a continuous fashion. Hence the Intermediate Value Theorem does not apply, and we can make no definitive statements concerning the question above.

For some interesting extra reading check out:

The intermediate value theorem is NOT obvious—and I am going to prove it to you, S.M. Walk, College Math Journal, September 2011.

2025-01-06 19:41:59