請問這題怎麼算

9. A rectangle is inscribed in a right triangle, as 000,00 shown in the accompanying figure. If the triangle 0 has sides of length 5, 12, and 13, what are the

請問這題怎麼算

1. What number exceeds its square by the largest amount? [Hint: Find the number x that maximizes trigin f(x) = x = x².] bas auoivdo líssi jon - sidan Enter Chaigh ot by

想請教這一題微積分，謝謝！

2. (8%) Consider a real numbers and suppose that lim x→0√√x+s4-s² find s (there are two possible answers). = 3,

想請問這題的(b)怎麼解！

Question 3: If f(x)=x³+3x+1, evaluate (a) f'(2) and (b) (-¹)(15). Sele

想請問紅色的地方要怎麼變藍色的

EXAMPLE 3 Solution We differentiate the equation implicitly. (AS) ² = x² + sin.xy d VIR SY d (²) - (x²)+(sin xy) = dx dx Find dy/dx if y² = x² + sinxy (Figure 3.31). d dx dy dx 2y = d 2x + (cos xy) (xy) dx ansa to along or see anuar Differentiate both sides with respect to x... ... treating y as a function of x and using the Chain Rule.

請問有人知道怎麼算嗎？拜託了🙏

11. [-/6 Points] DETAILS Find the constants a and b such that the function is continuous on the entire real line. a = b = LARCALC12M 1.4.073. 5, f(x) = ax + b, -5, 13 X ≤-2 -2<x<3 X23

微積分求解

8. [0/8 Points] Consider the following limit. DETAILS lim Ax-0+ lim Ax-0+ Submit Answer 8 X + Ax Ax Simplify the rational expression as much as possible. 8 x(x + Ax) X 8 X PREVIOUS ANSWERS X Evaluate the one-sided limit. (If an answer does not exist, enter DNE.) DNE LARCALC12M 1.4.029.EP.

請問這題要怎麼算～

1 89. Let f be a function such that f'(x): If 1 + x² 0₁5 2 LIFE AND Qurbliet 1 + x² dig(x) = f(2x + 1), what is g'(x)?sup inslissxs) =

想問第4題畫線處 分母(tan²x)-1 =(sin²x/cos²x)-1 =(sin²x-cos²x)/cos²x 為什麼把1/cos²x移到分子以後 分母不是sin²x-cos²x ? 感謝！

x² Example 4 Find lim x 0 sec X 1 SOLUTION The evaluation of this limit requires a little imagination. Since both the numerator and denominator tend to zero as x tends to zero, it is not clear what happens to the fraction. However, we can rewrite the fraction in a more amenable form by multiplying both numerator and denominator by sec x + 1. x2 x² sec x 1 lim x-0 secx 1 - = = x2 sec x 1 - x² (secx + 1) sec² x 1 lim (sin x-0 sin x sec x + 1 sec x + 1) - = Since each of these factors has a limit as x tends to 0, the fraction we began with has a limit: = x² (secx + 1) tan² x - 1 x² cos²x(secx + 1) sin² X 2 (*) ² (cos²x)(secx + 1). 2 ·lim cos²x lim (secx + 1) = (1)(1)(2) = 2. x→0 x ●

求解第24題

t (a), explain wh 2+x-6 x-2 lim (x + 3) imit, if it exists. 12. lim x² - 4x x-1 x²-3x - 4 14. lim 2x² + 3x + 1 2 x-1 X² 2x - 3 16. lim 18. lim h→0 20. lim X→-1 22. lim u-2 24. lim t-0 (2 + h) - 8 h 3x - 4 28. lim h→0 x² + 2x + 1 .4. - 1 26. lim x--4 (1 t 4u + 1 u - 2 3 t² + t x²+9-5 x + 4 1 (x + h)² h 1 X Illustrate tion of th 33. If 4x 34. If 2x - 35. Prove 36. Prove 37-42= explain 37. lim 39. 41. 43