微积分第四章课件

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.
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.
Since f (x) exists for all x, the only critical numbers of f occur
when f (x) 0, that is, x 0 or 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
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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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) Байду номын сангаас0 if x 3 , and f (x) does not exist 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).
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.
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,
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.
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;
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.
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
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.
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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
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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.
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]:
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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)
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.
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.
In particular, suppose that p(a) =p(b) =0,where a and b are real numbers and a≠b.
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.
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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,