微积分第四章课件

Since f (x) exists for all x, the only critical numbers of f occur
when f (x) 0, that is, x 0ห้องสมุดไป่ตู้or x 2. Notice that each of these
critical numbers lies in the interval (-1/ 2, 4). f (0) 1 f (2) 3 f ( 1) 1 f (4) 17 28
Chapter 4 Applications of Differentiation 4.1 Maximum and Minimum values Definition 1 A function f has an absolute maximum (or global maximum) at c if f (c) f (x) for all x in D, where D is the domain of f. The number f(c) is called the maximum value of f on D.
In particular, suppose that p(a) =p(b) =0,where a and b are real numbers and a≠b.
We conclude therefore that f’(d) =0.
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y
y f (x) Example 1 Lets apply
Rolle’s Theorem to the
position function s=f(t) of a
o ac
b x moving object. If the object
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Example 6 Find the absolute maximum and minimum
values of the function f (x) x3 3x2 1 1 x 4.
Solution
2 Since f is continuous on [-1/2, 4],We get
f (x) 3x2 6x 3x(x 2)
2
In Figure 1, the function takes on a local minimum at c. f(b) is a local maximum value of f(x); f(d) is a maximum value of f(x) on [a, e] and is also a local maximum value of f(x).
is in the same place at two
Figure 1
different instants t=a and t=b
then f(a)=f(b). Rolle’s Theorem says that there is
some instant of t=c between a and b when f’(c)=0;
f has an absolute minimum (or global minimum) at c if f (c) f (x) for all x in D, where D is the domain of f. The number f(c) is called the minimum value of f on D. The maximum and minimum values of f are called the extreme values of f.
that is, the velocity is 0. (In particular, you can see
that this is true when a ball is thrown directly
upward.)
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Example 2 Prove that the polynomial function p(x) =2x3+5x –1 has exactly one real zero.
when x 0. Thus, the critical numbers are 3 and 0. 2
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The Closed Interval Method To find the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]:
Figure 2
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Theorem 3 (Fermat’s Theorem) If f(x) has a local extreme value (that is,
maximum or minimum) at c and if f’(c) exists, then f’(c)=0.
c
d
Figure 3
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This means that c or d must lie in the open interval (a, b) and therefore f’(c) or f’(d) exists. Assume that f(x) takes on maximum value at d and d∈(a, b). We have f’(d) =f-’(d) =f+’(d). By the definition of the derivative,
f (x) 3x4 16x3 18x2 1 x 4 is shown in Figure 2. We can see that f(1)=5 is a local maximum, whereas the absolute maximum is f(-1)=37. Also, f(0)=0 is a local minimum and f(3)=-27 is both a local and an absolute minimum.
0 0
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Example 3 If f(x)=x3 , then f’(x)=3x2, so f’(0)=0. But f has no maximum or minimum at 0. Example 4 The function f (x) x has its (local and absolute) minimum value at 0, but that value can’t be found by setting f’(x)=0 because f’(0) does not exist. Definition 4 A critical number ( or point) 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.
Comparing these four numbers, we see that the absolute
maximum value is f(4)=17 and the absolute minimum value is
f(2)=-3.
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4.2 The Mean Value Theorem Rolle’s Theorem Let f be a function satisfying the following three hypotheses: 1 f is continuous on a 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 f’(c)=0. PROOF If f(x) is constantly k on [a, b], the result is obvious. If f(x) is not constantly k on [a, b], f(x) must take on a maximum value at some point d of [a, b] and a minimum value at some point c of [a, b]. Since f(a) =f(b), c or d cannot be a and cannot be b.
Note We can’t expect to locate extreme values simply by setting f’(x) =0 and solving for x.
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Proof For the sake of definiteness, suppose that f has a local maximum at c. Then,
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Definition 2 Let c be a number in the domain of a function f. 1 f(c) is a local (or relative) maximum value of f if there is an open interval I containing c such that f(c)≥f(x) for all x in I. 2 f(c) is a local (or relative) minimum value of f if there is an open interval I containing c such that f(c)≤f(x) for all x in I.
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Example 5 Find the critical numbers of f (x) x3/5(4 x).
Solution We have
f
( x)
3 5
x2 / 5
(4
x)
x3/5
(1)
12 8x 5x2/5
Therefore, f (x) 0 if x 3 , and f (x) does not exist 2
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 [a, b]
3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.
Proof Obviously, p(x) is continuous on [0, 1] and p(0) = -1<0, p(1) =6>0, we know that p(x) has
at least one real zero. Suppose that p(x) has more than one real zero.
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Example 1 If f(x)=x2 , then f(x)≥f(0) for all x. Therefore, f(0)=0 is the absolute (and local) minimum value of f. However, this function has no maximum value. Example 2 The graph of the function
y
a0b c
de x
Figure 1
f(a) is a minimum value of f(x) on [a, e], but it is not a
local minimum value of f(x) because there is no open interval I contained in [a, e] such that f(a) is the least value of f(x) on I.