Question

1. For each function in question 1 of section 6.1 exercises, now use second- order conditions to determine whether each stationary value you found is a maximum, minimum, or point of inflection.

Answer 1

Solution: ca y = x3–37*+12 for maxima cod minima cor) point of inflection, 3) 342 - 3(2x) =0 - 312-6950 6x = 372 and y. so :.x =0,2 Now at y =0 6x6 - at x 20 = 6(0)-6 afx=2 , = 662) -6 1. = 6. (tve) x=0 is point of maxima and x=2 is point of minimal

9 : 3-891 tqxn-hx = h » Slayd = (4x2-(4)(3)(x2)+16) -o > 4x3-12x2+160 = 23-342+4=0 -> CX+1) (x24x+4)=0 » (x+1) (x-2)(x-2) =0 - x = 2,2,-) Now did = 122°-24% = 12-12-24(+1) at x =-), - 12 +24 = 36 (tve) at 1=2 = 12(2)?-24(2) =1264) —24(2) x=-1 is the point of minima and X =2 is point of inflection of dx² co

(C) Y = 343–3x-2: → som et = 30)(2)-3=0 -) 9x²=3 - x=25 Now he = 188 at -18/8 (-ve) is point of max! X = t is point of minima and x = 1 9=3x9-10 X3 + 6x +1 ► Mei - 3 (443) – 10(322) + ((2x) =0 => 1247-30x2 +124 =0 → 1222_304 +120, x=0,

1942 - 30% +12=0 2x²_5x +2=0 2x²_ ux-x+2=0 21(x-2)-(x-2)=0 (x-2)(20-1) 30 x=0, x=2,0 = 1/2 : 24 C221201 +12 day = 19 (312) – 60(2x) +12 = 3642 – 12.00 +12 at X 30, dry = 360-12060) +12 =12 ( tve) at X = 2, they = 36(4) –12002) +12 = 144-240+12 = -84 (-ve) = 36(1/4) – 120/2 +12 - 9-60 +12 = -39 C-ve) x zo is point of minima and x=2, 1/2 point of max. - o - i at x = 1/2