求解第二題！ （因為圖片要兩題一起看所以順便拍進來了）

§ 1.1 An Intuitive Introduction to Limits For Exercises 1 and 2, use the graph of the function f to determine whether each statement is true or false. Explain.(從圖形決定極限) 1. O lim f(x)=XX/ x→-3+ lim f(x) = 1. x→2 2. a) lim f(x)=1. x-→-3+ e) lim f(x) does not exist. x →4+ c) lim f(x) = 2. x->2 b) lim x->0 flim f(x) = 3. x-5 d) lim f(x) = 3. x-4 f(x) = 2. Xe) lim f(x) does not exist. x →3 lim f(x) = 2. X→4 S b) lim f(x) = f(0). x->>0 d) lim f(x)=3. x->2* y 4 3 234 y = f(x)

求解釋這一題 看不太懂😭

例題:【反函數】 若f(x) = x²,f:[0,3]→[09],求 f(x)的反函數。 2: To9]→{0.3] 解: f(x) = x 在 [0,3]上為1-1 函數,取f'(x) = x,則有 f(f'(x)) = (√x)} = =x, ∀x∈[0,1] 345 S' (f(x))=√x² = =x,∀x∈[0,0],故f(x) 的反函數為」。 f(x)=xx 2:57/9:52 y X

自動控制 求解

Design a compensator such that the damping ratio < = 0.5 and the undamped natural frequency Wn 2 rad/sec. R(s) + G.(s) Compensator (a) Dominant closed-loop poles? (b) Angle deficiency? (show the diagram) (c) Lead or lag compensator? 1 0.1s+1 1 52 G(s) → C(s)

想問上面那題，為什麼區間是0、10？ 從哪裡得出來的

1-6-8 中間值及勘根定理 片 1 例題:【中間值定理】 給函數f(x) = x+x+1,證明必定存在一實數c,使得f(c)=100。 解: 因 f(x) = x+x+1為一個多項式,所以在[0,10]區間上連續 因為f(x) = x* + x+1在閉區間[0,10]上連續, 又(0)=1<100<1001 = f(10) 所以由中間值定理,必定存在一實數c,使得f (c)=100。 ↳s 例題:【勘根定理】 證明方程式x-2x²+x+1=0,在(-1,1)區間內至少有一個實根。 證:

想請問這題是如何解出來的？ 手寫的圖是我嘗試解的過程，但沒有解出答案，麻煩大家了🫠

21. A metal ion in a weak-field octahedral complex has two more unpaired electrons than the same ion does in a strong-field octahedral complex. Which of the following could the metal ion be? (A) Cr³+ (B)Mn²+ (C)Co²+ (D)Ti²+ (E)Cu²+

求救微積分神人，我完全看不懂這是什麼啊！！！

用導數定義求極限 f'(a) = limf(x) = f(a) x-a 解 解 8.lim √x+1+√√x - I x-0 √√x+1-√√x+1 x-a lim √x+1+ [(√x+1-1) + (√√x-1-1)]/(x-1) x-0 √√x+1+√√√√x+1 x-0 [√√x+1-1) +(√√x+1-1)]/(x-1) g(x)=√√x+1, g'(0)=- h(x)=√√x-1 h'(0) = 59. lim x-1* lim x-1 w√x - wx √√x - √x k(x)=√√√x+1, k'(0) = "√x - √/x_ √√x - √x 3 練習 3.4E 9 r+1 在解特殊型式之極限問題時有其用處。 + ·*. lim √x+1+√√√x −1 2 3 x-0√√√x+1-3√x+1 1 1 2 3 9 x-1 1 m 1 P 1 2₁√√x+1/x = 0 x-1 lim x-1² √√√x – 1 x-1 1 n 9 1 3/(x-1)² Jx=0 1 3√(x + 1)² %r+ "x-1_√x-1 = 1 = 2₁ X- Jx=0 x-1 √x-1 - 1 pq mn = = 5 m -(n-1) 1 1 3 1 -Xm m 1 -XP P 1 1 + lim n Xn 1 -X q 1 √x=1 √x+1= √1-x 3/1-X !

第一張是題目，第二張是答案，請大家看一下我哪裡寫錯了，一直跟答案對不上，如果有人知道如何解，請讓我知道如何求得答案，謝謝

1.1 模擬測驗 環境化學 Environmental Chemistry 9克的純硫酸(H2SO4)加水至0.5升,試求此硫酸溶液的重量百分濃度?體積莫 9x10² 耳濃度?含多少ppm?當量濃度? 0.5 500

請問這題為什麼只能先對y積分？只是因為先對x的話積不出來嗎？

Evaluate the double integral: f(x, y) = ye; R is bounded by x=2 2 and y=0. ud hisborod【107 東吳巨資】 解:積分區域R如圖,只能先對 y 積分 s! Si yet" dx dy - Syex! + dy 3X² So ye. 314² dy 4 = site dy A R x = | 2x y=2x [[ ye²d4 = ['ƒ”ª" ye^¹ dydx = ['-—- y^²e²" d dx R |y=0 y 4 y=2x 28²e²dx=3e²-3(e-1). = S₁ 2x² e ³ dx = 2²/ #gesidgate enyi lo-te 3 == X YOY'S nll > x = 1 1:2

請問這題求dy/dx的時候，為什麼不能用隱含數微分法來微？

例1 Find the area of the surface generated by revolving the curve 6xy=x+3 from | |x=1 to x=3 about the x-axis. 【100 成大】 x++3x² 解:6.xy = x + 3 > y= dv 1 1+(2)2 =₁/1+ 1+-x dx 4 所求表面積為 = 1 + 6x 6 2x 1 + 24x² 1 1 2 dy 1 -- dx 2 1 +--- 2x² x+ = 1 2x² 1 ·x² +· 2 2x²

想問這兩個式子怎麼知道什麼時候用加的什麼時候用減的啊？

Sidizzo son ei ji povaworl grey line tolera L apapagoslingeri 14.0 V gool sign S qisning nanalintuniendizzo 00 6.00 (+ 21.21. Snus sur le mure on Roomet while keeping it electrically the 4.0 2 es or parallel combinations of 918 b + a oissant 10.0 V sociated with combining resis-mollastunquodh (a) emoved and replaced by a wire qoo circuit.) Because the circuit is his problem is one in which we → I www 6.02 100 9.00 (2 18.00 bus M 2.0 2 abcda: (2) 10.0 V – (6.0 N) Į — (2.0 ₪) I₂ = 0 befcb: - (4.0 N) I2 − 14.0 V + (6.0 M) — 10.0 V = 0 - 1₂ Figure 21.21 (Example 21.4) led in branches. moq as labeled in Figure 21.21.mais12A circuit containing different i svig 19 d od oni bi u bliud tot lo mo battib 210109 26919rw 1+ en slin es (1) 1₁ + 1₂2-1₂ = 01S sugit ni nobı aldı YOU ARE C 15usdwes (2.02) 4₂2 01 Toyons de = L d (3) -24.0 V + (6.0 N) - (4.0 N)₂ = 0lvom snigul grans di anioq gnilise or of am gids de stod bad i mi smo duo d Monosh 10.0 V (6.02), − (2.0 N) (1,₁ + 1₂) = 0 Tom Tom (4) 10.0 V - (8.02) 4₁ 10 0 sob