Maximum Point on a set A Definition. Let f be a function and A a set of numbers contained in the domain of f. A point x in A is a maximum point for f on A if
f(x) ³ f(y) foreveryyinA.
The number f(x) itself is called the maximum value of f on A.
N.B.
Several texts are inconsistent in distinguishing the value, x, the value
of the function, f(x), and the point, (x,f(x)), on the
graph.
The basic relation between the maximum point for f on an
open interval and the derivative is given in Theorem 1.
Theorem 1. Let f be any function defined on (a,b). If x is a maximum point for f on (a,b), and f is differentiable at x, then f¢(x) = 0.
Definition. Let f be a function and A a set of numbers contained in the domain of f. A point x in A is a local [relative] maximum point for f on A if there is some d > 0 such that x is a maximum point for f on A Ç(x - d, x + d).
f(x) ³ f(y) foreveryyinA Ç(x -d,x+ d).
A less technical statement is that f(x) ³ f(y) for all
nearby points y in A.
Definition. A critical point of a function f is a number x such that
f¢(x) = 0.
The number f(x) is called a critical value of f.
N.B. Once again there is often inconsistency in
referring to x, f(x), and the point (x,f(x)) on the
graph.
To locate the maximum point of f on a closed interval
[a,b], we need only look at
·
critical points of f in [a,b] (usually a small number),
·
end points a and b,
·
points x in [a,b] such that f is
not differentiable (which should be obvious).
Rolle's Theorem. If f is continuous on [a,b] and differentiable on (a,b), and f(a) = f(b), then there is an x in (a,b) such that f¢(x) = 0.
Proof. If f is constant on [a,b], then
f¢(x) = 0 for all x in (a,b). If f is not
constant on [a,b], then there is a maximum point or
minimum point x for f on (a,b). At such a point, by
Theorem 1, f¢(x) = 0.
Applying Rolle's Theorem to various functions, we obtain several
important results.
Mean Value Theorem. If f is continuous on [a,b] and differentiable on (a,b), , then there is an x in (a,b) such that
f¢(x) =
f(b) - f(a)
b - a
.
.
Proof. Let
g(x)
= f(a) +
f(b) - f(a)
b - a
(x - a),
the secant line through (a,f(a)) and (b,f(b)).
Then
F(x)
= f(x) - g(x)
satisfies the hypotheses of Rolle's Theorem. There is an x in
(a,b) such that
F¢(x)
= f¢(x) -
f(b) - f(a)
b - a
= 0.
We could not resist the proof of a version of L'Hôpital's Rule.
Cauchy's Mean Value Theorem. If f and g are continuous on [a,b], and differentiable on (a,b), then there is an x in (a,b) such that
Proof. Without loss of generality, assume that
f(a) = g(a) = 0. Using (*), notice that there is a d > 0
such that, for 0 < |x - a| < d, g¢(x) ¹ 0
for 0 < |x - a| < d. By the Mean Value Theorem, for 0 < |x - a| < d, g(x) ¹ 0. Fix x. By the Cauchy Mean Value Theorem, there is a
cx (which depends on x) between a and x such that
f(x)g¢(cx) = g(x) f¢(cx),
f(x)g¢(cx)
= g(x)f¢(cx)
(f)
Dividing (f) by g(x) ( ¹ 0!) and g¢(cx), we
obtain
f(x)
g(x)
=
f¢(cx)
g¢(cx)
.
As x ® a, cx ® a, so that
lim
x ® a
f(x)
g(x)
=
lim
x ® a
f¢(cx)
g¢(cx)
.
=
lim
x ® a
f¢(x)
g¢(x)
.
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