单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,Chapter3 Applications of Derivatives,Chapter3 Applications of Deri,1,3.1 Extreme Values of Functions,3.1 Extreme Values of Function,2,3.1 Maximum and minimum values,3.1 Maximum and minimum values,3,The maximum and minimum values of,f,are called the extreme values of,f.,c,d,(c,f(c),(d,f(d),The maximum and minimum values,4,微积分英文ppt课件-1,5,Example :,Example :,6,Example:,Example:,Example:,Example:Example:Example:,7,We have seen that some functions have extreme,values, whereas others do not. The following theorem,gives conditions under which a function is guaranteed,to possess extreme values.,We have seen that some functio,8,Example,This function has no maximum or minimum,机动 目录 上页 下页 返回 结束,Example,This function has no maximum or minimum,Caution:,The conditions cannot be weakened.,ExampleThis function has no ma,9,Example,Caution:,The conditions is sufficient, but not necessary.,ExampleCaution:,10,This function has minimum value f(-1)=-4,This function has maximum value f(-4),This function has minimum val,11,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.,We start by looking for local extreme values.,Problem:,12,c,c,13,Proof:,Without loss of generality, we consider local,Minimum:,Proof: Without loss of general,14,Caution:,1) The condition is just sufficient. We cannot expect to locate extreme values simply by setting and solving for,x.,For example:,2)There may be an extreme value even when does not exist.,For example:,Caution:,15,Fermats Theorem does suggest that we should at least start looking for extreme values of,f,at the numbers c where or does not exist.,Definition:,A critical number of a function,f,is a number,c,in the domain of,f,such that either or does not exist.,Fermats Theorem does suggest,16,Conclusion:,If,f,has a local maximum or minimum at c, then c is a critical number of,f,.,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,or it occurs at an endpoint of the interval,.,Conclusion: If f has a local,17,The Closed Interval Method,To find the absolute maximum and minimum values of a continous function,f,on a closed interval a, b:,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.,The Closed Interval Method,18,Example,Find the absolute maximum and minimum values of the function,Solution:,Since f is continuous on the given closed interval, we can use the Closed Interval Method:,Example Find the absolute maxi,19,微积分英文ppt课件-1,20,微积分英文ppt课件-1,21,EXERCISE,EXERCISE,22,Solution:,Solution:,23,4.2 The Mean Value Theorem,Rolle s Theorem:,Let,f,be a function that satisfy the following three hypotheses:,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,b,a,c,4.2 The Mean Value TheoremRoll,24,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:,Case1:,M=m,means that,f (x),is a constant function, then ,so any number,c,in (a, b) is OK!,a,b,Proof Because f (x) is con,25,Case2:,mM,and,f(a)=f(b),means that the absolute maximum,M,and the absolute minimum,m,cant 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 Fermats theorem, we know that,c,Case2: mM and f(a)=f(b),26,X,Y,-1,1,0,Caution:,The conditions cannot be weakened.,Example,XY-110Caution:The conditions c,27,Example,X,Y,1,0,ExampleXY10,28,Example,Caution:,The conditions is sufficient, but not necessary.,ExampleCaution:,29,Example 2,Prove that the equation has exactly one real root.,Solution:,Let ,then,f (x),is continuous and,By the Intermediate Value Theorem, there is at least one number,c,such that,f (c) =0,.,On the other hand ,suppose that it had two roots, and f (x) is continuous on a,b and differentiable on (a, b). By the Rolle s theorem, we know that there is a number d in (a, b) such that,Example 2 Prove that the equa,30,But,That is a contradiction!,Therefore ,the equation cant have two real roots.,But,31,Figure 1,Example,Lets apply Rolles Theorem to the position function,s,=,f,(,t,) of a moving object. If the object is in the same place at two different instants,t=a,and,t=b,then,f,(,a,)=,f,(,b,). Rolles Theorem says that there is some instant of,t=c,between,a,and,b,when,f,(,c,)=0; that is, the velocity is 0. (,In particular, you can see that this is true when a ball is thrown directly upward.,),Figure 1Example Lets apply Rol,32,微积分英文ppt课件-1,33,微积分英文ppt课件-1,34,Example 4:,Example 4:,35,Analysis:,Analysis:,36,Let,be continuous on, differentiable on,. If, then for every real,there is at least one number,in,for which,.,Letbe continuous on, different,37,微积分英文ppt课件-1,38,微积分英文ppt课件-1,39,a,b,Geometric Interpretation:,x,abGeometric Interpretation:x,40,that is,According to Rolles Theorem ，,Such that,that is,so,Set,Proof:,is continuous on,It is obvious that,，,differentiable on,and,that is According to Rolle,41,Example,Lets consider,f,(,x,)=,x,3,-,x,a,=0,b,=2. since,f,is a polynomial, it is continuous and differentiable for all,x, so it is certainly continuous on 0, 2 and differentiable on (0, 2). Therefore, by the mean value Theorem, there is a number,c,in (0, 2) such that,f,(2)-,f,(0)=,f,(,c,)(2-0),so this equation becomes,6=(3,c,2,-1)2=6,c,2,-2,which gives,But,c,must lie in (0, 2), so,Figure 3,Example Lets consider f(x)=x,42,Example,Suppose that,f,(0)=-3 and,f,(,x,)5,for all values of,x,. How large can,f,(2) possibly be?,Solution Since,f,is differentiable (and therefore continuous) everywhere, we can apply the Mean Value Theorem on the interval 0, 2. There is a number,c,in (0, 2) such that,f,(2)-,f,(0)=,f,(,c,)(2-0),so,f,(2)=,f,(0)+2,f,(,c,)=-3 +2,f,(,c,),which gives,f,(2) -3+10=7. The large possible value for,f,(2) is 7.,Example Suppose that f(0)=-3,43,Example,Suppose f(x) is continuous on 0, 1 and differentiable on (0, 1). Prove that there is at least a number c in (0, 1) such that,f(1) = 2cf(c) + c,2,f(c).,Solution,We define a new function g(x) as follows:,g(x) = x,2,f(x),for every x in 0, 1.,Example Suppose f(x) is contin,44,Since f(x) is continuous on 0, 1 and differentiable on (0, 1), the same is true for the function g(x). So, the function g(x) satisfies the hypotheses of Lagranges Theorem and there exists at least a number c in (0, 1) such that,g(1) g(0) = g(c)(1 0),or, equivalently,1,2,f(1) 0,2,f(0) = 2cf(c) + c,2,f(c),i.e., f(1) = 2cf(c) + c,2,f(c).,Since f(x) is continuous on 0,45,Question:,Question:,46,Example,Prove that,Proof Let Since,f,(,t,) is continuous on 0,x, and differentiable on (0,x,), the function,f,(,t,) satisfies the hypotheses of Lagranges Theorem and there exists at least a number,c,in (0,x,) such that,Example Prove that,47,such that,or, equivalently,And,we have,such that,48,Example,If an object moves in a straight line with position function,s,=,f,(,t,) ,then the average,Velocity between,t=a,and,t=b is,and the velocity at,t=c is .,Example If an object moves in,49,Theorem:,If for all x in an interval ,then f is constant on .,Applications:,Theorem: If for a,50,微积分英文ppt课件-1,51,Corollary:,If for all,x,in an interval , then is constant on ;that is ,where,c,is a constant.,Proof :,Let .Then,Thus,where,c,is a constant.,for all,x,in,Corollary: If,52,Example,Proof that,Example Proof that,53,微积分英文ppt课件-1,54,