By Edgar Asplund; Lutz Bungart

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A) 4xz + 6 cos (2x - 3y + 4z) 4z) (d) (c) (b) 18 sin (2x - 3y + 4x - 24 cos (2x - 3y -l· 4z) 36 cos (2x - 3y + 4z) 4z) 49 (e) 4 + 4 8 sin (2x - 3y + T-1. |Ì (x) = il» f ( * + 0 - f < » ) 3U t 1 t- 0 = f(x); 3f ,YÌ . . f(x - t) - f(x) _ f(x - t) - f(x) — (x) - lim >—*- = - lim 3U Ü _t 2 t+ 0 t- 0 f(x+x) - f W - - f . ( x ) . = _llm τ τ + 0 f(X+t T-2. C > - f ( X ) . lim f(x - tu) - f(x) = _ lim f(x - tu) - f(x) t t -► 0 = _ t -y 0 _t f(X+xU)-f(X)=_|i(x)< l i m τ -> 0 T-5. 3, page 351 2. (a) 7 (b) 4 6.

T-10. Follows directly from Def. 2, and the definition of matrix multiplication. T-11. Take X = U - V in part (d) of Thm. 1 to obtain (a) and (b). For (c), simply observe that |-x| = |x|. For (d), let X = U - W and Y = W - V in Thm. 3. 6, page 210 2. λ2 - 2λ + 4; no eigenvalues 0 -3 4. -(λ + 1)(λ - 2 ) 2 ; λΊ = -1, 3 L 5J 33 λ 2 = λ 3 = 2 > 0 1 -2 -1 6. 5 ; λ2 = - 3 , 1 -λ(λ + 3)(λ - 4 ) ; λ1 = 0, L οJ L 2J 1 λ2 = 4, ν. 0 ^ 0 8. - ( λ - 1 ) ( λ 2 - 2λ + 2 ) ; λ = 1 , 1 Lo J ο 0 10. ; λ2 - 1 λχ = - 1 , ; λ3 = - 2 , i, 1J LOJ 12.

R=l The desired result is obtained by substituting this in the constraint equations. 1, page 435 2. (x, y, z) = {(x, y, z)|z φ 0, y φ ± 9} 59 y -9 (b) f^x, y) log (x - y ) , f 2 (x, y) = e (c) (d) f1(x,y,z) = 1 _ χ y y ; {(x, y)|x > y} * ( \ 1 , f2(x,y,z) = — 2» y - z ; {x,y,z)|x Φ y, x Φ z, y φ ±ζ}, f (x,y,z) = - ^ x-z χ^2 f^x, y) = , f 2 (x, y) = e 2 2 cos(x - y ) y 2z f 3 (x, y) / x 2 - y 2 ; {(x, y)||x| > |y|> 4. 8. (a) m = 3, n = 2 (b) m = n = 2 (d) m = 2, n = 3. 6. r~ "^ Γ 1 (a) (b) (c) (a) -2 0 9 0 2 5x + 5y 7u - v 5y 4u + 3v {(x,y,z)|x φ ±y} 6 2 (b) 5 5 7 -1 0 5 4 3 x + y + z 2u + 2v 10.