Finite Interval( 有限區間 ) open interval ( a, b) { a< < b} closed interval [ ab, ] { a b} hal open( or hal closed) interval ( ab, ] { a< b} [ ab, ) { a < b} Ininite Interval( 無限區間 ) [ a, ) { a < } (, b] { < b} ( a, ) { a< < } (, b) { < < b} (, ) { } < < R : closed interval : closed interval : open interval : open interval : both closed and open
De A unction( 函數 ) rom a set A to a set B is a rule that assigns to which element in A eactly one ( 恰有一個 )element y, called () in B A the domain( 定義域 ) o, denoted by D { } ( A) A the range ( 值域 ) o, denoted by R each input one and only one output y
[E] A { abc,, } B {,,3, 4} (a) : A B (b) g : A B a ( a) a b ( b) b 3 c 3 ( c) c 3 is a unction is an one to one unction ( 一對一函數 ) {,, } D A abc { } R A { ( a), ( b), ( c) } {,,3} B g is a unction g is not an one to one unction g g {,, } D A abc { } R g A { g( a), g( b), g( c) } {,3}
(c) h: A B a (not eactly one!) h is not a unction [E] (a) y + : R R y is an one to one unction D R R R y (b) ( 0) [ ) : 0, R y is not a unction since ± (not eactly one!)
Eercise: : y Is a unction when (a) + y (b) + y (c) + y (d) + y
Ans: (a) + y y : y Since y is unique (eactly one) or each is a unction. (b) + y y : y y Since is unique (eactly one) or each is a unction. (c) + y y ± : y Since y ± is not unique or each is not a unction. (d) + y y ± : y Since or each y ± is not a unction. is not unique
y + : R R y Since y +, we have y + or + { } So, D R R, R (the set o all allowable values o ) { y y } (, ) R R (the set o all possible unction values () )
[E] Find the domain( 定義域 ) and range ( 值域 )o the ollowing unctions (a) D { 0} (,0) ( 0, ) R {} - 0 R { y y 0} (,0) ( 0, ) R {} - 0
(b) D R { 0} [ 0, ) { y y 0} [ 0, ) (c) D R R 3 (, ) R (, )
(d) 3 (i) D : 3 0 3 (ii) R : { 3} [ 3, ) D 3 0 y 3 0 { y y 0 } [ 0, ) R (e) D R R { yy 0} [ 0, )
Eercise:. Find the domain( 定義域 ) and range( 值域 ) o the ollowing unctions (a) + (b) + (c) +. g. Find D g
[Sol]:. (a) D R + { } { y y 0} +. g 0 + 0 {, } D g [, ) (, ) [, ) - { } (b) + D R { } { y y } (c) D R { 0} {,} when > 0, when < 0,
Graphs o unctions( 函數的圖形 ) The graph o a unction () is the graph o the eq. y i.e. the set {( ) D },. [E] The piecewise deined unction( 分段定義函數 ) is given by, i, i > Sketch( 畫圖 )the graph. D R R [ 0, )
Graphs the absolute value unction( 絕對值函數圖形 ) [E] Graph the unction y 4 (i) irst graph y 4 (ii) then graph y 4
Eercise: () Graph the unction y using the skill you just learned. () Graph the piecewise deined unction( 分段定義函數 ):, <, (3) Find D and R when, <,
[Sol]: (), (), < (3) (i) D R (ii) when <, > 0 ie. > 0 when, ie. So R 0, { yy 0} >
Odd Function( 奇函數 ) Even Function( 偶函數 ) De: ( ) De: ( ) Graph: symmetric about the origin Graph: symmetric about the y-ais ( 對稱 ) ( 對稱 ) [E] 3 [E] 3 3 Since ( ) is an odd unction Since ( ) is an even unction symmetric about the origin symmetric about the y-ais
[E] Determine whether is even, odd or neither.( 都不是 ) 3 (a) (b) + 3 (c) + [Sol]: 3 3 ( ) (a) ( ) is an odd unction. ( ) (b) ( ) is an even unction. ( ) + + (c) ( ) ( ) + + 3 3 ( ) Since ( ) and ( ). is neither even nor odd. [E] A portion o the graph o a unction is given as ollows. (a) Complete the graph o i it is even. (b) Complete the graph o i it is odd. ( )
[Sol]: (a) i is even (b) i is odd. Eercises:. Determine whether is even, odd or neither. (a) (b) 3. The graph o a unction is given as ollows. Is it even, odd or neither?
[Sol]:. (a) ( ) is even (b) ( ) 3 is even. (a) The graph is symmetric about the orign ( 對稱原點 ) is odd (b) The graph is not symmetric about the orign nor the y-ais ( 不對稱原點與 y 軸 ) is neither odd nor even
The Vertical Line Test( 垂直線檢定 ) A curve C in the y-plane C is the graph o a unction o No vertical line intersects( 交叉 )the curve more than once. ( a) b & ( a) c a contradiction! ( 矛盾 ) This is a unction s graph This is not a unction s graph Eercises.5.6.7.8.3.4.5.6.7.8.3.3.34.35.39.4.43.44.45.48.5 57.58.59.60.6.6.63.65
. A Catalog o Essential Functions( 常用基本函數概述 ) () linear unction : y m+ b (its graph is a line 直線 ) ( 線性函數 ) () polynomial : ( 多項式 ) m : slope 斜率 b: y-intercept p a + a + + a + a + a ( a 0) n n n n 0 n (i) A polynomial o degree or 0 is a linear unction p a + a or p a 0 截距 n : nonnegative integer, called the degree ( 次數 ). a0, a,, an, an : constants, called the coeicients ( 係數 ). (ii) A polynomial o degree is a quadratic unction( 二次多項式 ) p a + b + c ( a 0) (its graph is a parabola) (iii) A polynomial o degree 3 is a cubic unction( 三次多項式 ) p a b c d a 3 + + + ( 0) ( 拋物線 )
(3) power unction( 冪函數 ) : (i) a n ; n: nonnegative integer n : polynomial a e: (a) (b) (c) 3 (d) (e) 4 5
(ii) when a ; n: positive integer( 正整數 ) n n root unction( 根式函數 ) e. 3 3 (a) (b) (iii) a. reciprocal unction ( 倒數函數 )
(4) rational unction ( 有理函數 ): P Q P and Q are polynomials + 3+ e., reciprocal unction 4 (5) algebraic unction ( 代數函數 ): a unction that can be constructed using algebraic operations (such as addition( 加 ), subtraction( 減 ), multiplication( 乘 ), division( 除 ) and taking roots( 取根式. 開根號 )) starting with polynomials. e. + (6) transcendental unction ( 超越函數 ): the unctions that are not algebraic. e. trigonometric unctions ( 三角函數 ): sin, cos, tan, inverse ( 反 ) trigonometric unctions eponential unctions logarithmic unctions + 3 : sin, cos, tan ( 指數函數 ): a ( a > 0, a ) ( 對數函數 ): log ( a > 0, a ) a,
[E] Classiy the ollowing unctions are one o the types o unctions that we have discussed. [Sol]: + 3 u t t + t + 5 (a) 5 (b) (c) 4 (d) 5 (e) () sin 5 (a) 5 : eponential unction. (transcendental unction) 5 (b) : power unction. (polynomial o degree 5) + (c) : algebraic unction. 4 (d) ut t+ 5 t : polynomial o degree 4. 3 (e) : rational unction. + () sin 5 : trigonometric unction. (transcendental unction)
(i) trigonometric unction (Appendi 附錄 D) π rad 360 or π rad 80 80 π So. rad 57.3 rad 0.07 rad. π 80 [E] (a) Find the radian( 弧度 ) measure o 60 5π (b) Epress rad in degrees. 4 [Sol]: π π (a) 60 60 rad rad 80 3 5π 5π 80 (b) rad 5 4 4 π Degrees 0 30 45 60 90 0 35 50 80 70 360 Radians 0 π π π π π 3π 5π 3π π 6 4 3 3 4 6 π
0 θ (hypotenuse) ( 斜邊 ) π (adjacent) ( 鄰邊 ) (opposite) ( 對邊 ) b sin θ cscθ c a cos θ secθ c b tan θ cotθ a c b c a a b y sin θ cscθ r cos θ secθ r y tan θ cotθ r y r y ( c osθ,sinθ ) (, + ) ( c osθ,sinθ ) (, ) ( c osθ,sinθ ) ( +, + ) ( c osθ,sinθ ) ( +, ) In calculus, we use radians( 弧度 ) to measure angles( 角度 ).
sine unction 正弦函數 cosine unction 餘弦函數 tangent unction 正切函數 cotangent unction 餘切函數 secant unction 正割函數 cosecant unction 餘割函數 y sin, < <, y y y cos, < <, y y sin π y tan, ( n+ ), < y < cos cos y cot, nπ, < y < sin π y sec, n+, y, y y cos y csc, nπ, y, y y sin ( + π ) T π co ( π ) ( ) tion) cos ( ) sin sin ( ) sin sin (odd unc s + cos ( T π) cos (even unction)
tan ( + π ) tan cot ( π ) + cot ( + π ) ( π ) sec sec csc + csc
.3 New Functions rom Old Functions (I) Transormations o unctions( 函數的變換 ) Vertical and Horizontal Shits( 垂直和水平平移 ) ( c > 0). y + c : shit the graph o y c units upward [E]. y c : shit the graph o y c units downward 3. y ( c) : shit the graph o y c units to the right 4. y ( + c) : shit the graph o y y + c units to the let 0, y (0) + 0, (0) +, y () +, () +, y () +, () + y 3 ( 0, y (0) 3 ( 0, (0) 3) ) y 0, y (0) 0, (0), y (), (), y (), ()
y ( + 3) 3, y (0), y (), y () 左移 3 單位 y 0, y (0), y (), y () 右移 單位 y ( ), y (0) 3, y () 4, y () [E] y + + 左移 單位上移 單位 y y 3 右移 3 單位下移 單位
[E] Sketch the graph o the unction + 6+ 0 ( 畫圖 ) [Sol]: The eq. o the graph is y + 6+ 0 ( ) y + 3 + () irst, sketch the graph () by shiting 3 units to the let and unit o the parabola y ( 拋物線 ) upward, we have the answer Eercise Use transormations to sketch the graph o y 3
Vertical Stretching ( 垂直延展 ) ( c > ). y c : stretch the graph o y vertically by a actor o c. y : compress the graph o y vertically by a actor o c ( 壓縮 ) c [E] y 0, y (0), y () y 0, y (0), y () y 0, y (0), y ()
Horizontal Stretching( 水平延展 ) ( c > ). y c : compress the graph o y horizontally by a actor o. y : stretch the graph o y c horizontally by a actor o c [E] c y 0, y 0, y () y () 0, y 0, y y 0, y 0 (), y
Relecting( 鏡射 ). y : relect the graph o about the -ais. y : relect the graph o about the y-as i [E] y ( 0, y ( 0) ) y ( 0, y ( 0) ) y (, y ( ) ) y ( ), y ( )
[E] Give the graph o y, use transormations to graphy, y +, y, y and y
[E] Sketch the graphs o the ollowing unctions: (a) y sin (b) y sin Notice: sin has period π Notice: sin sin ( ) Eercise Use transormations to graph y and y +
Algebra o Functions( 函數的代數 ) Let and g be unctions with domain A and B the sum ( 和 ) + g + g domain A B the dierence ( 差 ) g g domain A B the product ( 積 ) g g domain A B the quotient [E] [Sol] ( 商 ) domain { A B g 0} g g I and g 4, ind the unction + g, g, g and ( + g) + 4, D [ 0, ] ( g) 4, D [ 0, ] 3 ( g ) 4 4, D [ 0,] [ 0, ), D [, ] ( 4 0 4 ) D g g 4 4, D [ 0, ) + g g g g g 0 and 4 0 0 and ± 0 <
composite unction( 合成函數 ) ( g) g In general, g g [E] I + and g, ind each unction and its domain. [Sol] (a) g (b) g (c) (d) g g ( ) (a) g g + g { 0} { } [, ) D ( ) (b) g g g + + g { 0} { or } [, ) (, ] D ( ) (c) + + + D R ( ) (d) g g g g g g g Notice: D g R { 0 and 0} { and } { } D
.5 Eponential Functions( 指數函數 ) D R a ( a > 0) (, ) ( 0, ) Laws o Eponents a, b: positive number( 正數 ) ;, y: real number( 實數 ).. 3. 4. a a a a + y y y a a y y y ( a ) a ( 3 ) 5 7 ( 3 3 3 ) 3 3 ( 5 ) ( ab) a b 5 3 3 3 0 ( ) 3 5 3 5
[E] Sketch the graph o the unction y 3 and determine its domain ( 畫圖 ) and range. [Sol]: Domain R Range,3 > 0 < 0 y 3 < 3 Eercise Graph the unction 3 y and state the domain and range.
The number e The slope( 斜率 ) o the tangent line ( 切線 ) to y e at 0, is. [E4] Graph y e and state the domain and range. 4 e > 000000!! Domain R Range (, ) e > 0 e > 0 y e >
.6 Inverse Functions and Logarithms( 反函數與對數 ) De A unction is called an one-to-one unctio( 一對一函數 ) i whenever that is, it never takes on the same value twice. 3 is not one-to-one No horizontal line intersect the graph more than once. is one-to-one. Horizontal Line Test( 水平線檢定 ) A unction is one-to-one no horizontal line intersect its graph more than once 3 One can easily check is one-to-one whereas g isn't.
De Let be an one-to-one unction with domain A and range B Then its inverse unction has domain B and range A and is deined by y y or any y B Notice : y y D R (domain o range o ) R D (range o domain o ) [E3] ( ) [Sol]: I 5, 3 7 and 8 0, ind 7, 5 and 0. ( ) 7 3 3 7, 5 5 0 8 8 0
Cancellation equations( 消去方程式 ): ( ) ( ) or every A or every B 3 3 For,, we have 3 3 3 3 3 3 ( ), the cancellation equations satisied. Notice: 3 3 3 Three steps or inding the inverse unction o an one-to-one unction. Write y. Solve this equation or in terms o y 3. Interchange( 互換 ) and y, the resulting eq. is y 3 [E4] Find the inverse unction o + 3 [Sol]:. Write y +. Solve or in terms o y : 3 y 3. Interchange and y, we have Notice: 3 3 + 3 y 3. Thereore
[E] [Sol ]:. Does () have an inverse unction? Suppose eists, then ( ) 4 and 4 4 and 4 ( () is not an one-to-one unction.), a contradiction. Thereore we know doesn t eist, i.e. () doesn t have an inverse unction. We conclude that only an one-to-one unction has an inverse unction. [Sol ]: I we try to ind the inverse unction.. Write y. Solve or : ± y 3. Interchange and y, we have y ± this is not a unction. Hence, () doesn t have an inverse unction. Notice: i is not one-to-one doesn't have an inverse unction.
[E]. Find the inverse unction. [Sol]:. Write y. Solve or in terms o y : Notice: 3. Interchange and y, we have y y i.e. y [ ) Thereore, and D R 0, R [ ) The unction g,, is not one-to-one, but h, 0, is. Thereore, does not have an inverse unction, but does and g h h g( ) is not one-to-one h is one-to n -o e
( ab, ) is on the graph o ( ba, ) is on the graph o The graph o is obtained by relecting the graph o about the line y 3 3 () 8 0 0 3 () 8 0 8 0
[E5] Sketch the graphs o and its inverse unction using the [Sol]: same coordinate aes( 座標軸 ). y y y To ind the :. y y. 3. y y Eercise () Find the inverse unction o () Sketch the graphs o and its inverse unction using the same coordinate aes. 3 5
Logarithmic Functions( 對數函數 ) a ( a > 0, a ) is one-to-one log (the logarithmic unction with base a) y y a y log y a a ( 0, ) { 0} D R > R (, ) D R 3, log 3 8 8 log 8 3 log 0 ( ) ( ) and range (, ) e. g log has domain, Cancellation equations: ( ) ( ) log a loga a a e. log 5 0 5 log 0 ab, is on the graph o y a ba, is on the graph o y log The graph o y log is obtained by relecting the graph o y a about y a a
Laws o Logarithms :, y: positive numbers ( 法則 ) () log a( y) log a + log a y () log a log a log a y y r (3) log rlog a a log log Change o Base Formula : b log ( b > 0, b ) ( 換底公式 ) a b a e. log constant log log 5 log 5 log 5
Natural Logarithm( 自然對數 ) log e ln : natural logarithm ln y y e ( ) ( ) ln has domain 0, and range, Cancellation equations: lne ln e e ln e. ln e ( + ) e ln 3 + 3 ab, is on the graph o y e ba, is on the graph o y ln The graph o y ln is obtained by relecting the graph o y e about y
ln y ln + ln y ln ln ln y y ln r rln log a ln ln a ln ln a [E7] Find i ln 5 [Sol]: ln 5 [E8] Solve( 求解 )the equation [Sol]: 5 3 ln e ln0 5 3 e 0 e e ln 5 5 3 ln0 e [E9] Epress( 表式 ) ln 5 a + ln b 3 5 ln0 5 ln0 3 as a single logarithm. [Sol]: ln a+ ln b ln a+ ln b ln a+ ln b ln a b
[E] The mass o that remains rom a 4-mg sample ater t years is [Sol]: 90 Sr t 5 () m t 4. Find the inverse o this unction. () m t t 5 4 is one-to-one. t m eists. t 5 Solve the eq. m 4 or t: t m 5 4 t m 5 ln ln the time required or the mass to 4 reduce to 5 mg is t ln ln m ln 4 5 5 t ( 5) ( ln4 ln5) ln t ln m ln 4 56.8 years 5 ln 5 t ( ln 4 ln m) ln 5 i.e. ( m) ( ln 4 ln m) ln The unction gives the time required or the mass to decay to m milligrams.
[E] Sketch the graph o the unction [Sol]: y ( ) ln Eercise () Solve the eq. () Find the inverse unction o ( ) 3 ln ln + 3 (3) Sketch the graphs o and its inverse unction using the same coordinate aes.
Inverse Trigonometric Functions( 反三角函數 ) arcsine unction (inverse sine unction) ( 反正弦函數 ) sin y sin y π π y, arccosine unction (inverse cosine unction) ( 反餘弦函數 ) cos y cos y 0 y π,
sin? Let sin y sin y π 5π y,, 6 6 π π but y π hence y sin 6 cos? 0 cos cos π 3 π sin? 3 Let sin 3 θ 3 π π sin θ and θ hence θ cos? π 3 π sin 3 3 π 0 cos π cos
arctangent unction (inverse tangent unction) ( 反正切函數 ) tan y tan y π π < y <, < < y csc csc y π 3π, 0 < y or π < y y sec sec y π 3π, 0 y < or π y < y cot cot y < <, 0< y < π π π π () 4 π ( ) 3 π ( ) 4 < tan < tan tan 3 tan () cot cot sec sec csc csc π 4 3π 4 π 3 4π 3 π 6 7π 6 ( 0 cot < < π ) π < 3π π sec < 0 sec or π < 3π π < csc 0 csc or
π π sin ( sin ) or sin ( sin ) or cos cos or 0 (QuadrantⅠ, Ⅳ) ( 象限 ) π (QuadrantⅠ, Ⅱ) cos ( cos ) or π π tan ( tan ) or < < (QuadrantⅠ, Ⅳ) tan ( tan ) or all cot ( cot ) or 0 < < π (QuadrantⅠ, Ⅱ) cot ( cot ) or all π 3π sec ( sec ) or 0 <, π < (QuadrantⅠ, Ⅲ) sec( sec ) or or π 3π csc ( csc ) or 0 <, π < (QuadrantⅠ, Ⅲ) csc( csc ) or or
[E3] (b) Evaluate ( 求值 ) [Sol] sec 3 : tan sin 3 Suppose sin θ 3 then sinθ 3 hence tan sin 3 cos : 4 3 tan cos tan sec 3 sin sec 3 3 7 4 3 3 7 4 4 sin cos 3
sin : 7 tan sin 7 3 5 3 5 7 7 cos sin tan : 3 cos tan 3 3 0 3 csc tan 0
[E4] Simpliy the epression cos( tan ) 化簡 表示式 [Sol ] (Solve geometrically)( 幾何解法 ) cos tan + [Sol ] (Solve algebraically)( 代數解法 ) Let y tan (we're going to ind cos y) tan y sec y + tan y sec y ± + tan y π π But < <, we have sec > 0, thereore sec + tan cos( tan ) cos y sec y + tan y + y y y y
[E] Simpliy sin ( cos ) [Sol ] sin θ sinθ cosθ sin cos sin cos cos cos [Sol ] Let cos θ (we're going to ind sin θ ) cosθ Since sin θ + cos θ, sinθ ± cos θ But 0, we have sin 0, hence sin cos θ π θ θ θ sin cos θ sin θ sinθ cosθ cos θ cos θ
[E] Evaluate [Sol]: 3π 4π 7 3 () sin sin () cos cos 5π (3) tan tan (4) tan tan 4 π 3π π 3π 3π 7 7 7 π () sin sin () For 0, cos cos 4π π π 3 3 3 4π π OR: cos cos cos 3 3 sin ( s ) [ π ] cos cos cos cos, since 0, (3) tan tan π π < < 5π π π 4 4 4 5π π () 4 4 (4) Since tan tan or tan tan tan tan OR: tan tan tan π in or π
Eercises:. Evaluate () sin () sin 3 3 (3) cos (4) cos ( ) (5) tan 3 (6) tan 3 (7) sec (8) sec 3 3. Evaluate () tan sin () cos( tan 5) 4 3 (3) sin cos (4) cot sec 3
3. Evaluate 4π () sin sin () sin sin 3 3 π 4 (3) cos cos (4) cos cos 3 5 3π 4 (5) tan tan (6) tan tan 0 5π (7) sec sec (8) sec sec 6 ( e)
[Sol]: π π π 5π π π π 7π. () ()- (3) (4) (5) (6)- (7) (8) 4 4 6 6 3 3 6 6 5. () () (3) (4) 5 6 3 5 π π 4 π 7π 3. ()- () (3) (4) (5)- (6)0 (7) (8)e 3 3 3 5 4 6