考试样卷及参考答案

考试样卷及参考答案中南财经政法大学20 –20 学年第一学期期末考试试卷课程名称：《微积分》（A ）卷课程代号：_____________ 考试形式：闭卷、笔试 使用对象：留学生得分 一、看图作答,每题10分，共50分1 函数f 的图像如图所示（The graph of function f is given ） (a) Estimate the value of ()2f .(估计()2f ) (b) Estimate the values of x such that ()3f x =.（估计x ，使得()3f x =）(c) State the domain of f . （指出f 的定义域） (d) State the range of f .（指出f 的值域）(e) On what interval is f increasing?（f 的递增区间） ( f ) Is f one-to-one? Explain.（f 是一对一的吗？为什么）2 函数f 的图像如图所示，请在以下坐标系中画出所求函数的图像（The graph of is given, draw the graphs of the following functions ）(1)(1)1y f x =+- (2) ()1y f x =--3、利用下图求各函数值(Use the graph to evaluate each expression)4、The graph of f is given.(函数f 如图所示) (1) Find each limit or explain why it does not exist.（求下列极限或者说明极限为什么不存在）(a) ()3lim x f x -→- (b) ()3lim x f x +→-(c) ()3lim x f x →- (d) ()2lim x f x -→(e) ()2lim x f x +→ (f) ()2lim x f x → (2) State the equations of the vertical asymptotes.（指出函数的垂直渐近线方程）5、The graph shown gives the weight of a certain person as a function of age. Describe in words how this person ’s weight varies over time.What do you think happened when this person was 30 years old?得分 二、计算题,每题10分，共50分1、设()322(),31f x x x g x x =-=+，计算()()()()()()1,2,1,f f g fg f a g ⎛⎫+-- ⎪⎝⎭.(If ()322(),31f x x x g x x =-=+,find ()()()()()()1,2,1,f f g fg f a g ⎛⎫+-- ⎪⎝⎭)。

2、A function f has domain [-5,5] and a portion of its graph is shown. (a) Complete the graph of f if it is known that f is even.(b) Complete the graph of f if it is knownthat f is odd.3、求出下列函数的反函数(find a formula for the inverse of the functions)。

(1)43y x =- (2)4123x y x -=+ 4、设2()1,()3,()f x x g x x h x x =-==，求f h g h g f οοοο, 5、Evaluate the following limits(1) 236lim 3x x x x →-+-+ (2) 23lim 24t t t →-+20 —20 学年第一学期微积分卷标答一、看图作答1、（a ）2.8 (b) x=2.2 or x=5.6 (c) [-6,6] (d) [-4,4] (e) [-4,4] (f) 不是2、略3、（a ）4 (b) 3 (c) 0 (d) 不存在 (e) 4 (f ) -24、(a) -2 (b) 0 (c) 不存在 (d) -∞ (e) 3 (f) 不存在5、略二、计算题(1)()()14f g +=()()252fg =()112f g ⎛⎫-=- ⎪⎝⎭()32f a a a -=--(2) 略 (3) (1) 43x y -=(2) 3142x y x+=- (4) 231f g h x =-o o ()231g h f x =-o o (5) (1) -5 (2) 7中南财经政法大学20 –20 学年第二学期期末考试试卷课程名称：《微积分（下）》（B）卷课程代号：_L2300120__考试形式：闭卷、笔试使用对象：留学生2、Use the graph to state the absolute and local maximum andminimum values of the function.得分一、看图作答,每题10分，共20分1、The graphs of f and g are given. Use them to evaluate each limit, if it exists.If the limit does not exist, explain why.(1) []2lim2()()→-xf xg x(2)1lim106()→-xf x(3)21()lim()xx f xg x→-⋅(4) 22lim43()→⎡⎤-⎣⎦xx f x(5) []lim()()xxf x g x→-1、(1) If ()=f x ()f x '（本题10分）(2) Find equations of the tangent lines to the curve ()=f x at the points()1,2--and ()4,12.(1)2234lim 254→+∞-+-+x x x x x (2)lim x (3) 236lim 3→-+-+x x x x (4) 2416lim 4→--x x x3、 The derivative of a function f at a number a , denoted by ()f a ',is()()()limh f a h f a f a h→+-'=If this limit exists. Use the definition to evaluate (1)f 'and ()f a 'if the function is 2()5=--f x x x .（本题15分）4、Calculate ()f x 'of each following functions （本题25分，每小题5分） (1) 3()=⋅x f x x e (2) ()34()21=-+f x x x(3) 2()3=-f x x x(4) 321+=x y2、Evaluate the following limits （本题20分，每小题5分）(5) 22()31=+-x f x x x5 Find the absolute maximum and absolute minimumvalues of on the given interval.20 -20学年第二学期Calculus 期末考试试卷标答一 1、（1）4（2）2 （3）不存在（4）10 （5）02、absolute maximum ()44f =; absolute minimum ()70f = Local maximum ()()44,63f f ==;local minimum ()21f = 二1、(1) ()f x '=(2) At point ()-1,-2,()714f slope '-==,equation ()7214y x +=+ At point ()4,12,()1143f slope '==,equation ()111243y x -=-. 2、(1) 22343lim 2542x x x x x →+∞-+=-+(2) lim 2x = (3) 236lim53x x x x →-+-=-+ (4) 2416lim 84x x x →-=- 3、()()()20001155(1)(1)(1)lim lim lim 11h h h h h f h f f h h h→→→+-+-++-'===+= ()()()22200055()()2()lim limlim 21h h h a h a h a a f a h f a h ah hf a a hhh→→→+-+----+-+-'====-4、(1) 23()3x x f x x e x e '=⋅+⋅ (2) ()()243()32181f x x x x '=-+⋅-(3) 2()32f x x -'=+ (4) 2233202x x y y y y''+⋅=⇒=-(1) []3()31,0,3f x x x =-+ (2) []2(),0,21xf x x =+(5) ()()()()22222222312332()3131x x x x x x xf x xx xx ⋅+--⋅+-'==+-+-5、(1) ()[]22()333101,10,3f x x x x x '=-=-=⇒=±=∈()()()01,11,319f f f ==-=absolute maximum ()319f =,absolute minimum ()11f =- (2) ()()()[]2222221121()01,10,211x x xx f x x x xx⋅+-⋅-'===⇒=±=∈++()()()1200,1,225f f f ===absolute maximum ()225f =,absolute minimum ()00f =。