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 you 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 $f(c) = d$ .

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 are necessary.

Let’s see the Intermediate Value Theorem in action.

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

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

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. Compute $\begin {array}{c|c} x & f(x) \\ \hline 0.1 & \answer [given]{-1.869}\\ 0.2 & \answer [given]{-1.672}\\ 0.3 & \answer [given]{-1.403}\\ 0.4 & \answer [given]{-1.056}\\ 0.5 & \answer [given]{-0.625}\\ 0.6 & \answer [given]{-0.104}\\ 0.7 & \answer [given]{0.513} \end {array}$ By the Intermediate Value Theorem, $f$ has a root between $0.6$ and $0.7$ . Repeating the process $\begin {array}{c|c} x & f(x) \\ \hline 0.61 & \answer [given]{-0.046719}\\ 0.62 & \answer [given]{0.011528} \end {array}$ so by the Intermediate Value Theorem, $f$ 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 use to show that curves cross:

Explain why the functions $\begin{align*} f(x) &= \frac{1}{2}x\\ g(x) &= x\sin \left(\pi x\right)+1 \end{align*}$ intersect on the interval $\left [0,\frac {3}{2}\right ]$ .

To start, note that both $f$ and $g$ are continuous functions, and hence $h = f-g$ is also a continuous function. Now

$\begin{align*} h(0) &= f(0) - g(0) \\ &= \frac{1}{2}\cdot\answer[given]{0} - (\answer[given]{0}\cdot \sin(\pi\cdot\answer[given]{0})+1)\\ &= \answer[given]{-1}. \end{align*}$ and in a similar fashion $\begin{align*} h\left(\frac{3}{2}\right) &= f\left(\frac{3}{2}\right) - g\left(\frac{3}{2}\right) \\ &= \frac{1}{2}\cdot\answer[given]{\frac{3}{2}} - \left(\answer[given]{\frac{3}{2}}\cdot \sin\left(\pi\cdot\answer[given]{\frac{3}{2}}\right)+1\right)\\ &= \answer[given]{\frac{5}{4}}\\ \end{align*}$ Since $h(0)<0$ and $h\left (\frac {3}{2}\right )>0$ , there must be some point, $x=c$ , in between $x=0$ and $x=\frac {3}{2}$ where $h(c)=0$ . So we would have $f(c)-g(c)=0$ giving $f(c)=g(c)$ . Hence, $f$ and $g$ intersect on the interval $\left [0,\frac {3}{2}\right ]$ .

Now we return to a more subtle example of how the IVT can be utilized, which you already started investigating earlier:

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

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

To solve this problem, consider two functions:

$W_{\mathrm {Gabby}}(t) =$ the amount of water in Gabby’s bowl at time $t$ .
$W_{\mathrm {Dustin}}(t) =$ the amount of water in Dustin’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 $W_\mathrm {Gabby}(t_\mathrm {start})-W_{\mathrm {Dustin}}(t_\mathrm {start}) > 0$ and $W_\mathrm {Gabby}(t_\mathrm {finish})-W_{\mathrm {Dustin}}(t_\mathrm {finish}) < 0.$ Since the amount of water in a bowl at time $t$ is a continuous function, as water is ‘‘lapped’’ up in continuous amounts, $W_\mathrm {Gabby}-W_{\mathrm {Dustin}}$ is a continuous function, and hence the Intermediate Value Theorem applies. Since $W_\mathrm {Gabby}-W_{\mathrm {Dustin}}$ 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 $W_\mathrm {Gabby}(t_\mathrm {equal})- W_{\mathrm {Dustin}}(t_\mathrm {equal}) =0$ 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, Gabby and Dustin. Is there a time when Gabby and Dustin have the same amount of dry cat food in their bowls assuming:

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

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

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