微积分英文课件PPT (5)

The Extreme Theorem If f is continuous on a closed int erval [a,b],then f attains an absolute maxmum value f (c) and an absolute minimum value f (d ) at some numbers c and d in [a,b].
To find an absolute maximum or minimum of a continuous function on a closed interval , we note that either it is local {in which case it occur at a critical number by the above conclusion}
We have if
if
Therefore
Because Thus
exists,
Fermat’s Theorem does suggest that we should at least start looking for extreme values of f at the
numbers c where f (c) 0 or f (c) does not exist.
1) f is continuous on the closed interval [a,b]. 2) f is differentiable on the open interval (a,b). 3) f (a) = f (b) Then there is a number c in (a,b) such that
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Caution:
The conditions is sufficient, but not necessary.
Example
x2 2x 3
f
(
x)
x
2x 2
4 x 1 1 x 2 2 x3
This function has minimum value f(-1)=-4
Find the absolute maximum and minimum value
1
2
of the function f (x) x3 (1 x)3
x 1，2.
y' 1 3x 33 x2 (1 x)
解 Therefore, y 0 if 1 3x 0,
答
that is, x 1 , and y dose not exist 3
4.1 Maximum and minimum values
Definition A function f has an absolute maxiumu
(or global maximum) at c if f c f (x) for all x
in D, where D is the domain of f. The number of f (c) is called the maximum value of f on D. Similarly, f has an absolute minimum at c if f (c) f (x) for all x in D and the number f(c) is called the minmum valueof f on D. The maximum and minimum values of f are called the extreme values of f.
This function has maximum value f(-4)
Problem: The Extreme Value Theorem says that a
continuous function on a closed interval has a maximum value and minimum value. But it does not tell us how to find these extreme values.
Definition:
A critical number of a function f is a number c in the domain of f such that either f (c) 0 or f (c) does not exist.
Conclusion: If f has a local maximum or minimum at c, then c is a critical number of f.
when x 0, or x 1.
Thus , the critical numbers are x 0, x 1 and x 1
3
The values of yat these critical numbers and at
the endpoints of the interval are
x 1 0 1/3 1 2
解 f (x) 3 4 0 3 4 / 3 0 3 2
答 Comparing these numbers,we get
[kəm'pεə
ymax f (2) 3 2 ymin f (1) 3 4.
4.2 The Mean Value Theorem
Rolle’ s Theorem: Let f be a function that satisfy the following three hypotheses: hai'pɔθisi:z]
1) Find the values of f at the critical numbers of f
in (a,b).
2) Find the values of f at the endpoints of the interval.
3) The largest of the values from step 1) and step 2) is the absolute maximum value; the smallest of these values is the absolute minimum value.
We start by looking for local extreme values.
Fermat's Theorem If f has a local maximum or minmum at c , and if f (c) exists , then f (c) =0
c
Proof: Without loss of generality, we consider local Maximum:
f (c) 0
c
a
c
Caution: The conditions cannot be weakened.
Caution: The conditions cannot be weakened.
y
1
Example
This function has no
o
1x
maximum or minimum
y
Example
2
1
o 1 2x This function has no maximum or minimum
or it occurs at an endpoint of the interval.
The Closed Interval Method
To find the absolute maximum and minimum values of a continous function f on a closed interval [a, b]:
The best way to learn mathematics is to do mathematics
Chapter 4 Applications of differentiation
4.1 Maximum and minimum values 4.2 The Mean value Theorem 4.3 How Derivatives Affect the shape of a Graph 4.4 Indeterminate Forms and L’ Hospital’s Rule 4.5 Summary of Curve Sketching 4.6 Graphing with Calculus and Calculators 4.7 Optimization Problems 4.8 Applications to business and Economics 4.9 Newton’s Method 4.10 Antiderivatives
Case1: M=m means that f (x) is a constant function, then f (x) 0 ,so any number c in (a, b) is OK!
a
b
Case2: m<M and f(a)=f(b) means that the absolute maximum M and the absolute minimum m can’t be taken on at the endpoints at the same time. So ,without loss of generality, we suppose that M is taken on at some point c in (a,b), so f(c) =M is a local maximum and f(x) is differentiable at x=c, by the Fermat’s theorem, we know that
f (c) 0
a
c
b
Proof Because f (x) is continuous on [a, b], by the Extreme Value Theorem, we know that f (x) can take on its absolute maximum M and absolute minimum m there are two cases: