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反函数

Topic 1. When does 1f - exist ?

There are many criteria that can be checked to see if an inverse function exists. Some of them are :

● One-to-one: A function is called one-to-one if 12()()f x f x = implies that 1x =2x . This implies that every y -value of the function comes from a unique value of x . This condition is

equivalent to the geometric condition that every horizontal line meets the graph of ()y f x = in at most one point. This condition can be difficult to check, since you need to know that the entire graph looks like.

● Strict monotonicity: A function that is either increasing or decreasing on its domain is called

strictly monotonic . If a function is strictly monotonic on its domain, it has an inverse function. This condition is normally easier to check than the one-to-one condition, since we can find ()f x 'and check that either ()f x '> 0 for every value of x , or ()f x '< 0 for every value of x .

Definition 1. A function g is the inverse of the function f if 1. (())g f x x =for all x in the domain of f . 2. (())f g x x =for all x in the domain of g .

In this situation g is denote by 1f - and is called 〝f inverse.〞

The Key Fact on th Existence of Inverse

Theorem 1. f has an inverse f ?is one-to-one.

Exercises :

1.(a)Show that 7()21f x x x =++ has an inverse. Also prove that 7210x x ++= has exactly one real solution.

Proof : We know that 6()720,.f x x x '=+>?Hence the function is strictly increasing, which means that it has an inverse. Next , we show that 7210x x ++= has exactly one real solution.

Let 7()21f x x x =++ Then 7(1)(1)2(1)12f -=-+?-+=-and (0)1f =. Since 0 is between ?2 and 1, and f is continuous everywhere (becuase it’s a polynomial), the Intermediate Value Theorem guarantees that there is some number c in (?1, 0) for which ()0f c =. That is, the given equation has at least one root.

Assume otherwise that the equation has two distinct roots. Let a and b denote the two roots, so that ()()0f a f b ==. The function f is continuous and differentiable everywhere, so we can use the Mean Value Theorem to conclude that there is some number c in (a ,b ) such that ()()()()0f b f a f c b a '-=-=

This contradicts ()f x '> 0,x ?, so we must conclude that there are not two distinct roots. Thus there is at most one solution to the given equation.

2.Does 2()1f x x =- have an inverse function? If not, can we restrict the domain so that it does? Solution : Since the graph of 2()1f x x =- is a parabola, we know that it is not one-to-one. Therefore, it does not have an inverse. However, if we consider just 0x ≥ or 0x ≤, it will have an inverse.

Topic 2. How to find inverse function ?

Since we want to find a function that “undoes” ()f x , we can use that 1()x f y -= if and only if

()y f x =.

Steps for finding a formula for an inverse function: 1.Solve ()y f x = for x in terms of y . 2. Switch the x and y variables. 3. Set 1()f x y -=.

Remark :

https://www.doczj.com/doc/1412309542.html,ing y instead of ()f x in the first step is only used to make the notation easier when solving the equation.)

2.If ()f x represents a function, then the notation 1()f x -, read “f f inverse of x ”, will be used to denote the inverse of ()f x . Similarly, the notation 1()g x -, read “g inverse of x ”, will be used to denote the inverse of ()g x .

3.11

()()

f x f x -≠

. It is very important not to confuse function notation with negative exponents.

Exercises :

1.Let f be a function given by the rule 2

()5

x f x x +=+. (a)Given the rule for the function 1f -.

(b)What is the domain of 1f - ? Give your answer in interval notation. (c)What is the range of f ? Give your answer in interval notation.

(a)First , we filp x and y : 2

y x y +=

+. Solve for y : 12552(1)25,()1

xy x y y x x y f x x --+=+?-=-==-. Or 2()5x y f x x +==

+, 15252(1)25,()1

xy y x x y y x f y y --+=+?-=-∴==-. Hence we get 125().1

x f x x --=

- (b)The domain is (,1)(1,)-∞?∞. (All but 1x =, which makes denominator 0.) (c)The range of f is the domain of 1f - ∴ran f =(,1)(1,)-∞?∞.

2.Consider the function f :(,1)(0,1)(0,)-∞-?→∞given by the rule 2()f x x =. (a)Sketch the graph of f .

(b) Is f one -to -one ? (You may simply answer yes or no.) (c)Sketch the graph of 1f -.

(d)What is 1f -? In other words, define the function 1f -.

(e) Show that (1f -?f )(x ) =x for all appropriate x , and also that (f ? 1f -)(x ) =x for all appropriate x .

In each case, make sure to write down what the appropriate x 's are.

(a) (b) Yes!

().1

x f x x -<<=>??

(e) First we will check that (1f -?f )(x ) =x . For 01x << we have

112(())()||f f x f x x --====x

where the second equality follows because 201x << for these x and the final equality follows because 0x <.

For 1x -∞<<- we have

112(())()||()f f x f x x x --===-=--=x

where the second equality follows from the definition of 1f - because 21x > and the second to last equality follows since 0x <. This establishes 1f -?()f x x = for all x in the domain of f .

Next we will check that (f ?1f -)()x =x . For (0,1)x ∈ we have 12(())f f x x -== because

0x >. Finally, for (1,)x ∈∞ we have 12(())(f f x x -==again this is valid because 10x >>.

3.Consider the function 1

()1

x x e y f x e +==-.

(a)What is the domain of f ?

(b) Compute the inverse of 1f -.

Solution : (a) We must exclude the point where 1x e -= 0, i.e. where 0x =. So the domain is ： (?∞, 0)?(0, ∞).

(b)11

(1)1(1)111

x x x x x x e y y e y e e y y e e y ++=?-=+?-=+?=--

∴11()ln()1y x f y y -+==-, therefore 11

()ln()1

x f x x -+=-. (c)The range of f is the domain of 1f -. The domain of 1f - is the set of points where

x x +->0. Implies that (1)(1)01,1x x x x +->?<->. So the domain of 1f - is (,1)(1,)-∞-?∞. Thus the range of f is (,1)(1,)-∞-?∞.

Remark :

If we want to graph 1()y f x -= and have information about the graph of ()y f x =, we can graph

1()y f x -= fairly easily. To do this, note that if (,)x y is a point on the graph of ()y f x =, then

(,)y x is a point on the graph of 1()f x -. This means that we can reect the graph of ()y f x =about

the line y x = and get the graph of 1()y f x -=.

Topic 3. Derivatives of inverse functions

If we know ()f x ', we want to be able to easily compute 1()()f y -' for ()y f x =.

Theorem 1 (the Inverse Function Theorem )

If f is differentiable and strictly monotonic on an interval I , then 1f - is differentiable at the corresponding point ()y f x = and 1()()f y -'=1

()

f x '. This can also be written

1dx dy dy

=. Proof ：∵1(())f f x x -=, 1()(())()1f f x f x -''?=(Chain rule).

11()(())()f f x f x -'?=

', or 11

()()()

f y f x -'='.

Exercises :

1.Let 3()1y f x x x ==++. Find 1()(3)f -'.

Solution ：Note that 3y = corresponds to x =1. Also 2()31f x x '=+, so 1()(3)f -'=11

(1)4

f ='.

2.(a)Explain why sinh is a one-to-one function.

(b)Find a formula for the derivative of the inverse hyperbolic sine function 1sinh y x -=.

(c)Show that 1sinh ln(x x -=+.

(d)Use the result of part (c) to find the derivative of 1sinh x -. Compare with your answer to part (b).

Solution ：(a)sinh is an increasing function on its domain, so it passes the Horizontal Line Test, and is one-to-one.

(sinh cosh 0,.

x x d e e x x x dx -+==>?) (b)Let 1sinh sinh y x x y -=?=, differentiating with respect to x , then 1cosh y y '=?

1cosh sinh 1)cosh y y y y '?=

==-= ∴

sinh d x dx -=

. (c) Let 1

sinh y x -=,21

sinh 22y y y y

e e e x y e

---===?2210y y e xe --=.

Hence y

e x ==.

Note that y e >0，(0,x

have y e x =+. Therefore 11()sinh ln(y f x x x --===+.

(d)

sinh ln((1d d x x dx dx -===

3.Let f be a continuous function on the interval [a , b ]. Assume that f is twice differentiable on the interval a x b <<, and that ()0f x '> and ()0f x ''> on this interval. Let g be the inverse function of f .

(a)Find an expression for the second derivative of g .

(b)Show that ()g x ''<0 on its interval of definition. Thus g is convex in the opposite direction of f .

Solution ：(a)Since 1

()(())

g x f g x '=

', the chain rule and the rule for differentiating quotients apply,

leading to the following expression for g '':

(())()

()((()))

f g x g x g x f g x '''''=-

'. (b)For a < x < b we have ()0f x '> and ()0f x ''>. Therefore we see from the above expression of (),g x ''that ()0g x ''< on its interval of definition.

4.Let ()f x '=. Find a formula for 1()()f x -', given that f is one-to-one and its derivative satisfies the indicated function.

Solution ：1(())f f x x -= , by chain rule 11

()(())()f f x f x -'='

. Thus 1()(())f f x -'=. Therefore we have 1()()f x -'

5.(a)Let 43()1,02,f x x x x =++≤≤and 1()()g x f x -=, find (3)g '. (b)If ()(7())F x f g x =, then (3)?F '=

Solution ：(a)∵1()()g x f x -=, ∴(())g f x x =,1(())()

g f x f x '='. Put 43()311f x x x x ==++?=, then 321111

(3)(1)43|7

x g f x x ='=

=='+. (b)Since ()(7())F x f g x =, chain rule applies. ()(7())7()F x f g x g x '''=??,(3)1g =.

Thus 1(3)(7)7(3)1519715197

F f g '''=??=??=.

Topic 4. A formula for integrating inverse functions

If the function ()y f x =has the integral ()f x dx ?, its inverse function 1()x f y -=can be easily integrated with the formula

111()()(())()()f y dy yf y yd f y f x x f x dx ---=-=?-???. Exercises :

1.Find

Solution:11sin xd x --=-=+?

=1x x C -++

2.Suppose f is continuous , f (0)=0, f (1)=1,)(x f '>0, and .3

)( 1

=?dx x f Find the value of the integral 11 0

() .f x dx -? Solution: 1111110 00 () ()|(())x f x dx xf x xd f x ---==-??1110

1(())(())f f x d f x --=-? 1

0211()133

f x dx =-=-

=?.

反函数定义

反函数定义一般地，设函数y=f(x)(x∈A)的值域是C，根据这个函数中x,y 的关系，用y 把x表示出，得到x= g(y). 若对于y在C中的任何一个值，通过x= g(y)，x在A 中都有唯一的值和它对应，那么，x= g(y)就表示y是自变量，x是因变量y的函数，这样的函数x= g(y)(y∈C)叫做函数y=f(x)(x∈A)的反函数，记作y=f^-1(x). 反函数y=f^-1(x)的定义域、值域分别是函数y=f(x)的值域、定义域. 反函数性质（1）互为反函数的两个函数的图象关于直线y=x对称；函数及其反函数的图形关于直线y=x对称（2）函数存在反函数的充要条件是，函数的定义域与值域是一一映射；（3）一个函数与它的反函数在相应区间上单调性一致；（4）大部分偶函数不存在反函数（唯一有反函数的偶函数是f(x)=a^x,x∈{0}，但是y=k（常数）无法通过水平线测试，所以没有反函数。）。奇函数不一定存在反函数。被与y轴垂直的直线截时能过2个及以上点即没有反函数。若一个奇函数存在反函数，则它的反函数也是奇函数。 (5)一切隐函数具有反函数；

(6)一段连续的函数的单调性在对应区间内具有一致性；（7）严格增（减）的函数一定有严格增（减）的反函数【反函数存在定理】。（8）反函数是相互的且具有唯一性（9）定义域、值域相反对应法则互逆（三反） (10)原函数一旦确定，反函数即确定（三定）(在有反函数的情况下，即满足（2））例：y=2x-1的反函数是y=0.5x+0.5 y=2^x的反函数是y=log2 x 例题：求函数3x-2的反函数解：y=3x-2的定义域为R，值域为R. 由y=3x-2解得 x=1/3(y+2) 将x,y互换,则所求y=3x-2的反函数是 y=1/3(x+2)（x属于R）（11）反函数的导数关系：如果X=F(Y）在区间I上单调，可导，且F‘（Y）不等于0，那么他的反函数Y=F’（X）在区间S={X|X=F(Y),Y属于I }内也可导，且[F‘（X)]'=1\[F’（Y）]'。反函数说明 ⑴在函数x=f’(y)中，y是自变量，x是函数，但习惯上，我们一般用x表示自变量，用y 表示函数，为此我们常常对调函数x=f‘(y)中的字母x,y，把它改写成y=f’(x)，今后凡无特别说明，函数y=f(x)的反函数都采用这种经过改写的形式。

解析几何专题训练理科用

解析几何专项训练班级学号成绩（一）填空题 1、若直线m my x m y mx 21=++=+与平行，则m =_-1____． 2、若直线2+=kx y 与抛物线x y 42 =仅有一个公共点，则实数=k 1 ,02 3、若直线l 的一个法向量为()2,1n =，则直线l 的倾斜角为 arctan2π- （用反三角函数值表示） 4、已知抛物线2 0x my +=上的点到定点(0,4)和到定直线4y =-的距离相等，则 m = -16 5、已知圆C 过双曲线 116 92 2=-y x 的一个顶点和一个焦点，且圆心C 在此双曲线上，则圆心C 到双曲线中心的距离是 16 3 6、已知直线1l ：210x y +-=，另一条直线的一个方向向量为(1,3)d =，则直线1l 与2l 的夹角是 4 π 7、已知直线:0l ax by c ++=与圆1:2 2 =+y x O 相交于A 、B 两点，3||=AB ，则OA ·OB = 12 - 8、若直线m 被两平行线1:10l x y -+=与2:30l x y -+=所截得线段的长为22，则直线m 的倾斜角是 0015,75 . 9、若经过点(0,2)P 且以()1,d a =为方向向量的直线l 与双曲线132 2 =-y x 相交于不同两点A 、B ，则实数a 的取值围是 2215,3a a <≠ . 10、（理科）设曲线C 定义为到点)1,1(--和)1,1(距离之和为4的动点的轨迹．若将曲线

C 绕坐标原点逆时针旋转 45，则此时曲线C 的方程为__22 142 y x +=___________． 11、等腰ABC ?中，顶点为,A 且一腰上的中线长为3,则三角形ABC 的面积的最大值 2 12、如图，已知OAP ?的面积为S ，1OA AP ?=．设||(2)OA c c =≥，3 4 S c =，并且以O 为中心、A 为焦点的椭圆经过点P ．当||OP 取得最小值时，则此椭圆的方程为 22 1106 x y += . （二）选择题 13、“2a =”是“直线210x ay +-=与直线220ax y +-=平行”的（ B ）条件（A ）充要；（B ）充分不必要；（C ）必要不充分；（D ）既不充分也不必要 14、如果i +2是关于x 的实系数方程02 =++n mx x 的一个根，则圆锥曲线 12 2=+n y m x 的焦点坐标是（ D ）(A))0,1(±； (B))1,0(±； (C))0,3(± ；(D))3, 0(± 15、已知：圆C 的方程为0),(=y x f ，点),(00y x P 不在圆C 上，也不在圆C 的圆心上，方程0),(),(:'00=-y x f y x f C ，则下面判断正确的是……( B ) (A) 方程'C 表示的曲线不存在； (B) 方程'C 表示与C 同心且半径不同的圆； (C) 方程'C 表示与C 相交的圆； (D) 当点P 在圆C 外时，方程'C 表示与C 相离的圆。 16、若双曲线221112211:1(0,0)x y C a b a b -=>>和双曲线22 2222222 :1(0,0)x y C a b a b -=>>的焦点相同，且12a a >给出下列四个结论：①2222 1221a a b b -=-； ②1221 a b a b >； ③双曲线1C 与双曲线2C 一定没有公共点； ④2121b b a a +>+；其中所有正确的结论序号是（ B ）A. ①② B, ①③ C. ②③ D. ①④ y P x o A

高一数学反函数的概念

4.5反函数的概念一、教学内容分析 “反函数”是《高中代数》第一册的重要内容.这一节课与函数的基本概念有着紧密的联系，通过对这一节课的学习，既可以让学生接受、理解反函数的概念并学会反函数的求法，又可使学生加深对函数基本概念的理解，还为今后反三角函数的教学做好准备,起到承上启下的重要作用. 二、教学目标设计（1）理解反函数的概念，并能判定一个函数是否存在反函数；（2）掌握求反函数的基本步骤，并能理解原函数和反函数之间的内在联系；（3）通过反函数概念的引入；函数及其反函数图像特征的主动探索，初步学会自主地学习、独立地探究问题；掌握观察、比较、分析、归纳等数学试验研究的方法；体验探索中挫折的艰辛与成功的快乐，激发学习热情. 三、教学重点与难点：反函数的概念及求法；反函数的图像特征；反函数定义域的确定. 四、教学流程设计五、教学过程设计 1、设置情境，引出概念引例：在两种温度度量制摄氏度(C )和华氏度（F ）相互转化时会发现，有时两人选用相同的数据，如下表，所建立的函数关系和作出的图像完全不同，这是为什么呢？

教师点拨：指导学生观察上面两个函数的异同，引出反函数的定义.介绍反函数的记号 )(1 x f y ；了解)(1 x f 表示反函数的符号，1 f 表示对应法则. 2、探索研究，深化概念 ①探求反函数成立的条件. 例1（1）2 x y （R x ）的反函数是（2）2 x y （0 x ）的反函数是（3）2 x y （0 x ）的反函数是学生活动：讨论函数反函数成立的条件（理论根据为函数的定义）：对值域A 中任意一个y 值，在定义域D 中总有唯一确定的x 值与它对应，即x 与y 必须一一对应. ②探求求反函数的方法.（课本例题）例2．求下列函数的反函数：（1）24 x y （2）13 x y （3）)0(12 x x y （4）)2 1 ,(2413 x R x x x y [说明]：学生分四组完成，教师巡视，把典型错误及正确解法投影. 学生活动：探求求反函数的方法. （1）变形:解方程，)(x f y 得)(1 y f x ；（2）互换:互换y x ,的位置，得)(1 x f y ；（3）写出定义域:注明反函数的定义域. ③观察反函数的图像，探讨互为反函数的两个函数的关系.

高中数学选择、填空题专项训练(21-28与答案)

1 / 29 综合小测21 一、选择题 1．已知θ为三角形的一个内角，且θθθθcos sin ,21cos sin 22y x -=+则方程=1表示（） A ．焦点在x 轴上的椭圆 B ．焦在点y 轴上的椭圆 C ．焦点在x 轴上的双曲线 D ．焦点在y 轴上的双曲线 2．双曲线 116922=-y x 两焦点为F 1，F 2，点P 在双曲线上，直线PF 1，PF 2倾斜角之差为,3π则△PF 1F 2面积为（） A ．163 B ．323 C ．32 D ．42 3．要使直线)(1R k kx y ∈+=与焦点在x 轴上的椭圆172 2=+a y x 总有公共点，实数a 的取值范围是（） A ．10≤

2 / 29 7．a 、b 是两条异面直线，下列结论正确的是（） A ．过不在a 、b 上的任一点，可作一个平面与a 、b 都平行 B ．过不在a 、b 上的任一点，可作一条直线与a 、b 都相交 C ．过不在a 、b 上的任一点，可作一条直线与a 、b 都平行 D ．过a 可以且只可以作一个平面与b 平行 8．已知点F 1、F 2分别是双曲线122 22=-b y a x 的左、右焦点，过F 1且垂直于x 轴的直线与双曲线交于A 、B 两点，若△ABF 2为锐角三角形，则该双曲线的离心率e 的范围是 A ．),1(+∞ B ．)21,1(+ C ．)3,1( D ．)21,21(+- 9．过抛物线x y =2 的焦点F 的直线m 的倾斜角m ,4 π θ≥ 交抛物线于A 、B 两点，且A 点在x 轴上方，则|FA|的取值范围是（） A ．]2 21,41 (+ B ．)1,4 1[ C ．]1,4 1( D ．),2 1(+∞ 10．在正方体ABCD —A 1B 1C 1D 1中，O 为AC 、BD 的交点，则C 1O 与A 1D 所成的角为 A ．60° B ．90° C ．3 3arccos D ．6 3arccos 11．直平行六面体ABCD —A 1B 1C 1D 1的棱长均为2， 60=∠BAD ，则对角线A 1C 与侧面DCC 1D 1所成角的正弦值为（） A ． 2 1 B ． 2 3 C ． 2 2 D ． 4 3 12．正方体ABCD —A 1B 1C 1D 1中，P 在侧面BCC 1B 1及其边界上运动，且总保持AP ⊥BD 1，则动点P 的轨迹是（） A ．线段 B 1C B ．线段B C 1