함수의 연속

Let $f(x)=x^3$. Since $f$ is a polynomial, it is continuous everywhere. If $f(a)<7<f(b)$, then $\sqrt[3]{7}$ is somewhere between $a$ and $b$. If we can find $a$ and $b$ that satisfy these inequalities, and are very close together, that will give us a good approximation for $\sqrt[3]{7}$.

• Let's start with integers. $1^3<7<2^3$, so $\sqrt[3]{7}$ is in the interval $(1,2)$.
• Let's narrow this down, say by testing $f(1.5)$. $(1.5)^3 = 3.375<7$, so $\sqrt[3]{7}$ is in the interval $(1.5,2)$.
• Let's narrow further, say by testing $f(1.75)$. $(1.75)^3\approx 5.34<7$, so $\sqrt[3]{7}$ is in the interval $(1.75,2)$.
• Testing various points, we find $f(1.9)<7<f(2)$, so $\sqrt[3]{7}$ is between $1.9$ and $2$.
• By testing more, we find $f(1.91)<7<f(1.92)$, so $\sqrt[3]{7}$ is in $(1.91,1.92)$.
• In order to get an approximation for $\sqrt[3]{7}$ that is rounded to two decimal places, we have to know whether $\sqrt[3]{7}$ is greater or less than $1.915$; indeed $f(1.915) \approx 7.02>7$, so $\sqrt[3]{7}<1.915$; then rounded to two decimal places, $\sqrt[3]{7}\approx 1.91$.

If this seems like the obvious way to approximate $\sqrt[3]{7}$, that's good. The IVT is a formal statement of a very intuitive principle.