## Extreme Value Theorem

**If f is continuous on [a,b] then f has a maximum and a minimum on the
interval.**

By the Bounded Value
Theorem, f(x) is bounded on [a,b]. That means f has an upper bound,
and by completeness, f has a least
upper bound.

Let M be the least upper bound of f on the interval [a,b].

Assume f has no maximum on the interval, which means that f(x) < M for all
x in [a,b].

Then g(x) = 1 / (M-f(x)) is continuous on [a,b], which by the Bounded Value
Theorem, means g(x) has an upper bound, say, N.

N ≥ g(x) = 1 / (M-f(x)) for all x in [a,b],
so

0 < 1/N ≤ M-f(x), so

f(x) < f(x)+1/(2N) < f(x)+1/N ≤ M,
which means there is a number strictly between f(x) and M, contradicting the
choice of M as the *least* upper bound of f.

**How is this theorem used?**

Rolle's Theorem -- that if
f(a)=f(b) then f'(c)=0 for some c in (a,b) -- depends on two key facts: that
any function has a maximum on a closed interval (the Extreme Value Theorem),
and that Relative
Extrema Occur Only at Critical Numbers.

### Internet references

Source of this proof: Cut-the-knot

### Related pages in this website

Go back to Calculus Home

Upper Bound -- definition of
"upper bound" and "least upper bound" of either sets or
functions

Relative
Extrema Occur Only at Critical Numbers -- if c is an extremum then f'(c)=0
or f'(c) is undefined

Rolle's Theorem -- that if
f(a)=f(b) then f'(c)=0 for some c in (a,b)

The webmaster and author of this Math Help site is
Graeme McRae.