Math4121 Lecture 3

Continue on Differentiation

Mean Value Theorem

Theorem 5.9 Generalized Mean Value Theorem

If f,g:[a,b]\to \mathbb{R} are continuous on [a,b] and differentiable in (a,b), then there exists a point x\in (a,b) such that


[f(b)-f(a)]g'(x)=[g(b)-g(a)]f'(x)

Proof:

Define h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x), t\in [a,b].

We need to show that there exists a point x\in (a,b) such that h'(x)=0.

By previous theorem, it's enough to show that h has a local minimum or maximum in (a,b). By Extreme Value Theorem


\begin{aligned}
h(a)&=[f(b)-f(a)]g(a)-[g(b)-g(a)]f(a)\\
&=f(b)g(a)-f(a)g(b)\\
h(b)&=[f(b)-f(a)]g(b)-[g(b)-g(a)]f(b)\\
&=g(a)f(b)-g(b)f(a)
\end{aligned}

So h(a)=h(b).

Consider three cases:

h is constant on [a,b]. Then h'(x)=0 for all x\in (a,b).
\exists t\in (a,b) such that h(t)>h(a)=h(b). Since every continuous function on a compact interval attains its supremum, and h(t)>h(a)=h(b), the supremum of h on [a,b] is attained at some point x\in (a,b). (Apply Extreme Value Theorem to h on [a,b].)
\exists t\in (a,b) such that h(t)<h(a)=h(b). Since every continuous function on a compact interval attains its infimum, and h(t)<h(a)=h(b), the infimum of h on [a,b] is attained at some point x\in (a,b). (Apply Extreme Value Theorem to h on [a,b].)

In all cases, h has a local minimum or maximum in (a,b).

QED

Theorem 5.10 Mean Value Theorem

The mean value theorem is a special case of the generalized mean value theorem when g(x)=x (the identity function).

If f:[a,b]\to \mathbb{R} is continuous on [a,b] and differentiable on (a,b), then there exists a point x\in (a,b) such that


f(b)-f(a)=f'(x)(b-a)

Intermediate Value Theorem

Definition 5.12.1 Intermediate Value

We say that f:[a,b]\to \mathbb{R} attains the intermediate values if for each \lambda\in (f(a),f(b)) there exists a point x\in (a,b) such that f(x)=\lambda.

Theorem 5.12.2 Continuous Function attains Intermediate Values

If f:[a,b]\to \mathbb{R} is continuous on [a,b], then f attains every value between f(a) and f(b).

Theorem 5.12 Intermediate Value Theorem

If f:[a,b]\to \mathbb{R} is differentiable on [a,b]. Then f' attains intermediate values.

Proof:

Let \lambda\in (f'(a),f'(b)).

Let our auxiliary function be g(t)=f(t)-\lambda t.

Since g'(t)=f'(t)-\lambda, it suffices to find x\in (a,b) such that g'(x)=0.

g'(a)<0 and g'(b)>0.

We claim that \exists t_1\in (a,b) such that g(t_1)<g(a).

If this were false, then for all t\in (a,b) we would have g(t)\geq g(a).


\frac{g(t)-g(a)}{t-a}\geq 0\\

Continue on Monday.

2.9 KiB Raw Blame History