What is the difference between Intermediate Value Theorem and the Extreme Value Theorem?

Jun 29, 2015

The Intermediate Value Theorem (IVT) says functions that are continuous on an interval $\left[a , b\right]$ take on all (intermediate) values between their extremes. The Extreme Value Theorem (EVT) says functions that are continuous on $\left[a , b\right]$ attain their extreme values (high and low).

Explanation:

Here's a statement of the EVT: Let $f$ be continuous on $\left[a , b\right]$. Then there exist numbers $c , d \setminus \in \left[a , b\right]$ such that $f \left(c\right) \setminus \le q f \left(x\right) \setminus \le q f \left(d\right)$ for all $x \setminus \in \left[a , b\right]$. Stated another way, the "supremum " $M$ and "infimum " $m$ of the range $\setminus \left\{f \left(x\right) : x \setminus \in \left[a , b\right] \setminus\right\}$ exist (they're finite) and there exist numbers $c , d \setminus \in \left[a , b\right]$ such that $f \left(c\right) = m$ and $f \left(d\right) = M$.

Note that the function $f$ must be continuous on $\left[a , b\right]$ for the conclusion to hold. For example, if $f$ is a function such that $f \left(0\right) = 0.5$, $f \left(x\right) = x$ for $0 < x < 1$, and $f \left(1\right) = 0.5$, then $f$ attains no maximum or minimum value on $\left[0 , 1\right]$. (The supremum and infimum of the range exist (they're 1 and 0, respectively), but the function never attains (never equals) these values.)

Note also that the interval must be closed. The function $f \left(x\right) = x$ attains no maximum or minimum value on the open interval $\left(0 , 1\right)$. (Once again, the supremum and infimum of the range exist (they're 1 and 0, respectively), but the function never attains (never equals) these values.)

The function $f \left(x\right) = \frac{1}{x}$ also does not attain a maximum or minimum value on the open interval $\left(0 , 1\right)$. Moreover, the supremum of the range does not even exist as a finite number (it's "infinity").

Here's a statement of the IVT: Let $f$ be continuous on $\left[a , b\right]$ and suppose $f \left(a\right) \ne f \left(b\right)$. If $v$ is any number between $f \left(a\right)$ and $f \left(b\right)$, then there exists a number $c \setminus \in \left(a , b\right)$ such that $f \left(c\right) = v$. Moreover, if $v$ is a number between the supremum and infimum of the range $\left\{f \left(x\right) : x \setminus \in \left[a , b\right]\right\}$, then there exists a number $c \setminus \in \left[a , b\right]$ such that $f \left(c\right) = v$.

If you draw pictures of various discontinuous functions, it's pretty clear why $f$ needs to be continuous for the IVT to be true.