微积分第四章ppt课件

,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,*,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 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,.,f,has an,absolute minimum,(or,global minimum,) at,c,if 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,.,1,Chapter 4 Applications,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,.,2,Definition 2 Let c be a numbe,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,a 0 b c d e 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,.,3,In Figure 1, the function,Example 1,If,f,(,x,)=,x,2, 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,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.,Figure 2,4,Example 1 If f(x)=x2 , then f(,Theorem 3,(Fermats 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.,Note,We cant expect to locate extreme values simply by setting,f,(,x,) =0 and solving for,x,.,d,c,Figure 3,5,Theorem 3 (Fermats Theorem),Proof,For the sake of definiteness, suppose that,f,has a local maximum at,c,. Then,6,Proof For the sake of definit,Example 3,If,f,(,x,)=,x,3, then,f,(,x,)=3,x,2, so,f,(0)=0. But,f,has no maximum or minimum at 0.,Example 4,The function has its (local and absolute) minimum value at 0, but that value cant 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.,7,Example 3 If f(x)=x3 , then f,Example 5,Find the critical numbers of,Solution,We have,8,Example 5 Find the critical nu,The Closed Interval Method,To find the absolute maximum and minimum values of a continuous 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 ,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.,9,The Closed Interval Method To,Example 6,Find the absolute maximum and minimum values of the function,Solution,Since,f,is continuous on -1/2, 4,We get,Comparing these four numbers, we see that the absolute maximum value is,f,(4)=17 and the absolute minimum value is,f,(2)=-3.,10,Example 6 Find the absolute ma,4.2 The Mean Value Theorem,Rolles 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,.,11,4.2 The Mean Value Theorem 1,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,We conclude therefore that,f,(,d,) =0,.,12,This means that c or d must li,Figure 1,Example 1,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.,),13,Figure 1Example 1 Lets apply R,Example 2,Prove that the polynomial function,p,(,x,) =2,x,3,+5,x,1 has exactly one real zero.,Proof,Obviously,p,(,x,) is continuous on 0, 1 and,p,(0),= -,10, we know that,p,(,x,) has at least one real zero.,Suppose that,p,(,x,) has more than one real zero. In particular, suppose that,p,(,a,) =,p,(,b,) =0,where,a,and,b,are real numbers and,a,b,.,14,Example 2 Prove that the polyn,Without loss of generality, assume that,a,0,for all,x,(since,x,2,is nonnegative) so,p,(,x,) 0,for all,x,. Thus, we have a contradiction and we can conclude that,p,(,x,) has exactly one real zero.,15,Without loss of generality, as,(,The Mean Value TheoremLagranges,Theorem,),Let,f,be a function satisfying the following hypotheses:,1,f,is continuous on a closed interval ,a, b,;,2,f,is differentiable on the open interval (,a, b,);,Then there is a number,c,in (,a, b,) such that,f,(,c,),=,(,f,(,b,),-f,(,a,),/,(,b-a,),or, f,(,b,),-f,(,a,),=f,(,c,)(,b-a,),.,Figure 2,16,(The Mean Value TheoremLagran,PROOF,OF,THE MEAN-VALUE THEOREM,We can create a function,g,(,x,) that satisfies the conditions of Rolles Theorem. It is not hard to see that,is exactly such a function.,If,f,is differentiable on (,a,b,) and continuous on ,a,b, then so is,g,. As you can check,g,(,a,) and,g,(,b,) are both 0. By Rolles Theorem, there is at least one number,c,in (,a,b,) for which,g,(,c,) = 0. So, we obtain,17,PROOF OF THE MEAN-VALUE THEORE,Example 3,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 3,18,Example 3 Lets consider f(x)=,Example 4,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.,19,Example 4 Suppose that f(0)=-3,Example 5,Given that for all real number,x, show that for all real numbers,x,1,and,x,2,.,Solution,Let,since,f,(,x,) is continuous on ,x,1,x,2, and differentiable on (,x,1,x,2,), the function,f,(,x,) satisfies the hypotheses of Lagranges Theorem and there exists at least a number,c,in (,x,1,x,2,) such that,With we have,20,Example 5 Given that,Example 6,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,or, equivalently,And,we have,21,Example 6 Prove that21,Example 7,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),=,2,cf,(,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.,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),=,2,cf,(,c,),+ c,2,f,(,c),i.e., f,(1,) =,2,cf,(,c,),+ c,2,f,(,c,),.,22,Example 7 Suppose f(x) is co,Corollary 1,If,f,(,x,)=0 for all,x,in an interval (,a, b,),then,f,(,x,) is constant on (,a, b,).,Corollary 2,If,f,(,x,),=g,(,x,) for all,x,in an interval (,a, b,), then,f,(,x,),=g,(,x,),+c, where,c,is a constant.,23,Corollary 1 If f(x)=0 for al,24,24,Example 8,Proof that,25,Example 8 Proof that25,4.3 How Derivatives Affect the Shape of a Graph,Increasing/Decreasing Test,Suppose,f,(,x,) is,continuous on ,a,b, and differentiable on (,a,b,).,(1) If,f,(,x,)0,for all,x,in (,a,b,), then,f,(,x,) is strictly increasing on ,a,b,.,(2) If,f,(,x,)0,for all,x,in (,a,b,), then,f,(,x,) is strictly decreasing on ,a,b,.,PROOF,Choose any two points,x,1,and,x,2,in ,a, b, with,x,1,0 and the intervals on which,f,(,x,)0 or (3,x+,5)(,x,-1)0,we get,x,1.,29,Example 3 If the f(x)=x3+x2-,Thus the function,f,(,x,) is increasing on (-, -5/3 and on 1, +).,It is easy to check that,f,(,x,)0 if 5/3 ,x,0,for all,x,in,I, then the graph of,f,(,x,) is,concave,upward,on,I,.,If,f”,(,x,)0 when,x,-1/3 and,f”,(,x,)0 when,x-,1/3.,By Concavity Test, we get the following results: the graph of,f,(,x,),is concave downward on,(,-, -,1/3),and is concave upward on,(,-,1/3, +,),.,So the point,(-1/3, -88/27),is an,inflection point.,The graph of,f,is,sketched in Figure 2.,Figure 2,35,Solution Since,Definition,A point (,c, f,(,c,) on the graph of,f,(,x,) is called an,inflection point,if,f,(,x,) is continuous at,c,and the graph changes from,concave upward,to,concave downward,or from,concave downward,to,concave upward,at the point.,The Second Derivative Test,Suppose that,f,(,x,) is differentiable on an open interval containing,c,and that,f,(,c,)=0.,1,If,f”,(,c,),0, then,f,(,x,) has a,local minimum,at,c,.,36,Definition A point (c, f(c),Example 6,Discuss the curve,y=f,(,x,),=x,4,-,4,x,3,with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve.,Solution,To find the critical numbers we set,f,(,x,)=0 and obtain,x=,0 and,x,=3. To use the Second Derivative Test we evaluate,f”,at these critical numbers:,f”,(0)=0,f”,(3)=360,37,Example 6 Discuss the curve y,Since,f,(3)=0 and,f”,(3)0,f,(3)=-27 is a local minimum. Since,f”,(0)=0, the Second Derivative Test gives no information about the critical number 0. But since,f,(,x,)0 for,x,0 and also for 0,x,0,The points (0,0) and (2,-16) are the inflection points. Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in Figure 3.,Interval,Sign of,f”,+,-,+,Concavity,upward,down ward,upward,Figure 3,39,The points (0,0) and (2,-16) a,Example 7,Sketch the graph of the function,f,(,x,),=x,2/3,(6-,x,),1/3,.,Solution,Interval,(-, 0),(0, 4),(4, 6),(6, +),f,+,f,decreasing on (-, 0),increasing on (0, 4),decreasing on (4, 6),decreasing on (6, +),40,Example 7 Sketch the graph of,By the First derivative Test,f,(0)=0 is a local minimum and,f,(4)=2,5/3,is a local maximum. The sign of,f,does not change at 6, so there is no minimum or maximum there. Looking at the expression for,f”,(,x,), we have,f”,(,x,)0 for,x,0 and for 0,x,0 for,x,6. Thus f is concave downward on (-, 0) and (0, 6) and concave upward on (6, +), and the only inflection point is (6,0). The graph is sketched in Figure 4.,Figure 4,41,By the First derivative Test,4.4 Indeterminate Forms and LHospitals Rule,Suppose,f,and,g,are differentiable and,g,(,x,) 0 near,a,(except possibly at,a,). Suppose that,42,4.4 Indeterminate Forms and L,Example 1,Find,Solution,Since,Remark,It is important to understand that LHOSPITALS RULE does not apply to quotients in general; you should verify first that the numerator and denominator both tend to zero.,43,Example 1 Find43,For example,but a blind application of LHospitals rule would lead to,This is incorrect.,Example 2,Find,SOLUTION,44,For example,44,Note,LHospitals Rule is also valid for one sided limits and for limits at positive infinity or negative infinite; that is, “xa” can be replaced by any of the following symbols: “xa,+,”, “xa,-,”, “x+”, “x-”.,Example 3,Calculate,SOLUTION,45,Note LHospitals Rule is als,The Forms 0, 0,0, ,0, 1 and -,(1) If and , then the form is called an indeterminate form of type 0. We can deal with it by rewriting the product,f(x)g(x),as a quotient,and then applying LHospitals Rule to the resulting indeterminate form of type 0/0 or /.,46,46,(2) If and , then the form is called an indeterminate form of type -.To find out the limit, we convert the difference,f,(,x,),-g,(,x,) into a quotient or a product and then apply LHospitals Rule.,(3),47,47,Example 4,Find,Solution,The graph of the function is shown in Figure 5.,Figure 5,48,Example 4 FindThe graph of th,Example 5,Find,Solution,(1) The given limit is an indeterminate form of,type 0.We write,49,49,Q:,50,Q:50,4.5 Summary of Curve sketching,Guidelines for Sketching Curve,Domain E. Intervals of Increase or Decrease,B. Intercepts F. Local Extreme Values,C. Symmetry G. Concavity and Points of Inflection,D. Asymptotes H. Sketch the Curve,51,4.5 Summary of Curve sketchin,Example 1,Use the guidelines to sketch the curve,The domain is,B. The,x,- and,y,-intercepts are both 0.,C. Since,f,(-,x,)=,f,(,x,), the function,f,is even.,D.,Therefore, the line y=2 is a horizontal asymptote.,52,Example 1 Use the guidelines t,Therefore, the lines,x,=1 and,x,=-1 are vertical asymptotes.,E.,Since,f,(,x,)0 when,x,0 (,x,-,1) and,f,(,x,)0,f,is increasing on (- , -1) and (-1, 0) and decreasing on (0, 1) and (1, ).,F. The only critical number is,x,=0. Since,f,changes from positive to negative at 0,f,(0) is a local maximum.,53,Therefore, the lines x=1 and x,G.,We have,Thus, the curve is concave upward on the intervals,(- , -1) and (1, ) and concave downward on,(-1, 1). It has no point of inflection. We sketch the curve in Figure 1.,Figure 1,54,G.Figure 154,Slant (Oblique) Asymptotes,The line,y = ax +b,is a,slant asymptote,for a function,f,(,x,) if and only if,exist.,Example 2,Find the slant asymptotes, if any, of the function,Solution,The domain of,f,(,x,) is -,x,0 for all,x,c,and,f,(,x,),c, then,f,(,c,) is the global maximum value of,f,.,(b) If,f,(,x,)0 for all,x,0 for all,x,c, then,f,(,c,) is the global minimum value of,f,.,59,4.6 Optimization ProblemsFirs,Example 1,A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the