3.2. Extrema and the Mean Value Theorem http://www.ck12.org
Theorem:Iff(c)is an extreme value offfor some open interval ofc,and iff′(c)exists, thenf′(c) = 0.
Proof:The theorem states that if we have a local max or local min, and iff′(c)exists, then we must havef′(c) = 0.
Suppose thatfhas a local max atx=c.Then we havef(c)≥f(x)for some open interval(c−h,c+h)withh> 0.
Sof(c+h)−f(c)≤ 0.
Consider limh→ 0 +f(c+hh)−f(c).
Sincef(c+h)−f(c)≤0, we have limh→ 0 +f(c+hh)−f(c)≤limh→ 0 + 0 = 0.
Sincef′(c)exists, we havef′(c) =limh→ 0 f(c+hh)−f(c)=limh→ 0 +f(c+hh)−f(c), and sof′(c)≤ 0.
If we take the left-hand limit, we getf′(c) =limh→ 0 f(c+hh)−f(c)=limh→ 0 −f(c+hh)−f(c)≥ 0.
Hencef′(c)≥0 andf′(c)≤0 it must be thatf′(c) = 0.
Ifx=cis a local minimum, the same argument follows.
Definition
We will callx=cacritical valuein[a,b]iff′(c) =0 orf′(c)does not exist, or ifx=cis an endpoint of the
interval.
We can now state the Extreme Value Theorem.Extreme Value Theorem: If a functionf(x)is continuous in a closed interval[a,b], with the maximum offatx=c 1
and the minimum offatx=c 2 ,thenc 1 andc 2 are critical values off.
Proof:The proof follows from Fermat’s theorem and is left as an exercise for the student.
Example 1:
Let’s observe that the converse of the last theorem is not necessarily true: If we considerf(x) =x^3 and its graph,
then we see that whilef′( 0 ) =0 atx= 0 ,x=0 is not an extreme point of the function.
Rolle’s Theorem: Iffis continuous and differentiable on a closed interval[a,b]and iff(a) =f(b),thenfhas at
least one valuecin the open interval(a,b)such thatf′(c) =0.
The proof of Rolle’s Theorem can be found at http://en.wikipedia.org/wiki/Rolle’s_theorem.
Mean Value Theorem: Iffis a continuous function on a closed interval[a,b]and iff′contains the open interval
(a,b)in its domain, then there exists a numbercin the interval(a,b)such thatf(b)−f(a) = (b−a)f′(c).