Question

(a) For the limit lim (x3 + x + 2) = 4, use a graph to find the largest possible value of that corresponds to s = 0.3. (Round your X-1 answer down to three decimal places.) 8 = 0.071 (b) By using a computer algebra system to solve the cubic equation x3 + x + 2 = 4 + ε, find the largest possible value of that works for any given a > 0. 5(E) = x(&)-1 (C) Put & = 0.3 in your answer to part (b). (Round your answer down to three decimal places.) 6(0.3) = 0.071 How does 6(0.3) compare with your answer to part (a)? 5(0.3) is larger than the S from part (a). 8(0.3) is the same as the 8 from part (a). 8(0.3) is smaller than the from part (a).

Answer 1

Ang! (6) Given Cubic equation 3+*+2 = 4+E by Wolfram Alpha the Real solution is H(E) = 0+381571415, 1.73205081 2742 +108&tli2 +9€+18 Calen 0.873580465 for Partic 3/1.7320508) 1478410 1108€ +112 +9€ +18 M(0-3) = 1.07111688 S (0-3) 2 0.071 > S(E) = M(C) – (H) Given A water trough is 5m. long and We have to find how fast is the water level disting when the water is loom deep. Volume of the trough = Area of cross-section x Length of the trough V = Axl l= 5m. (Constant) Differentiate both side with respect to t dr = ex dA at Area of any trapezoid = length of upper base + lower base] 2 X height 50 40 30 40 To 30 m then length of upper base at any time is 2m +40 & 60 h → h= 2m " length of upper base at any time is 2m+40=ht40 ) Area of toapezoid (Ch+40) +40/xh (cross-section) (4 +40) Xh A = h2 +40h het Differentiate both sides with respect to t d Th [(a+h) + 40 xh = d (h2+40h) Xdh. SA da Ch+40) X dh dt put this value in ① dv = ex Ch+40) Xdh IF Given dv = 0.3 mitrin. = 0.3X106 cm2/min, lc 5mi = 500 (m. & h is required moment dt It is 10 cm. 0:3x106 = 500 X (10+40) Xdh = 5000 x 5Qxdh at 0.3 3X166 103 a. 300 25 - dh at dh = 12 cm. /min. | Ang dt flance The water level is increasing at the rate of 12 cm. ./min. - dh 12 0 m/min = 2 m./min. dh A = 0.12 m./min Ang Scanned with CamScanner cs Scanned with CamScanner