MthT 430 Notes Chapter 11 Significance of the Derivative

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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) ≥ f(y) for every y in A.

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 δ > 0 such that x is a maximum point for f on A ∩(x − δ, x + δ).

f(x) ≥ f(y) for every y in A ∩(x −δ, x + δ).

A less technical statement is that f(x) ≥ 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

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critical points of f in [a,b] (usually a small number),

•

end points a and b,

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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

[f(b) − f(a)]g′(x) = [g(b) − g(a)]f′(x).

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Proof. Apply Rolle's Theorem to

H(x) = [f(b) − f(a)](g(x) − g(a)) − [g(b) − g(a)](f(x) − f(a)).

Theorem 9 (L'ÔPITAL'S RULE). Suppose that

lim x → a  f(x) = 0, and lim x → a  g(x) = 0.

Suppose that

lim x → a  f′(x) g′(x) exists.

(*)

Then

lim x → a  f(x) g(x) exists,

and

lim x → a  f(x) g(x) = lim x → a  f′(x) g′(x) .

Proof. Without loss of generality, assume that f(a) = g(a) = 0. Using (*), notice that there is a δ > 0 such that, for 0 < |x − a| < δ, g^′(x) ≠ 0 for 0 < |x − a| < δ. By the Mean Value Theorem, for 0 < |x − a| < δ, g(x) ≠ 0. Fix x. By the Cauchy Mean Value Theorem, there is a c_x (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) (†)

Dividing (†) by g(x) ( ≠ 0!) and g^′(c_x), we obtain

f(x) g(x) = f′(cx) g′(cx) .

As x → a, c_x → 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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