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By Mazzone F.D., Cuenya H.H.

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Then we may write d d [f (x)g(x)h(x)] = [f (x)φ(x)] = f 0 (x)φ(x) + f (x)φ0 (x) dx dx = f 0 (x)g(x)h(x) + f (x) [g 0 (x)h(x) + g(x)h0 (x)] = f 0 (x)g(x)h(x) + f (x)g 0 (x)h(x) + f (x)g(x)h0 (x) 7. (a) 8. (a) x2 − 3 x2 (b) − 9 x2 (c) d (ax + b) = a dx 1 d −a (c) = dx ax + b (ax + b)2 30 (x + 5)2 (d) acx2 + 2adx − bc (cx + d)2 d x(ax + b) = 2ax + b dx d ax + b −b (d) = 2 dx x x (b) 9. (a) Yes; the continuity of f (x) is a necessary condition for f (x) to be differentiable. , Fig. 1c). 10. 3 1. dy/dx = (dy/du)(du/dx) = (3u2 + 2((−2x) = −2x[3(5 − x2 )2 + 2] 2.

The MC curve must be upward-sloping throughout. Since the increasing segment 4. If b=0, then the MC-minimizing output level becomes Q∗ = − of MC is associated with the convex segment of the C curve, b = 0 implies that the C curve will be convex throughout. 5. (a) The first assumption means π(0) < 0. Since π(0) = k, we need the restriction k < 0. (b) Strict concavity means π 00 (Q) < 0. Since π 00 (Q) = 2h, we should have h < 0. (c) The third assumption means π 0 (Q∗ ) = 0, or 2hQ∗ + j = 0. Since Q∗ = −j/2h, and since h < 0, the positivity of Q∗ requires that j > 0.

The first-order condition requires that e2x = 1 and 4y = 0. Thus x∗ = y ∗ = 0 so that z∗ = 4 2 Since fxx fyy = 4(4) exceeds fxy = 0, z ∗ = 4 is a minimum. 5. (a) And pair (x, y) other than (2, 3) yields a positive z value. (b) Yes. At x∗ = 2 and y ∗ = 3, we find fx = 4 (x − 2)3 and fy = 4 (y − 3)3 = 0 (c) No. At x∗ = 2 and y ∗ = 3, we have fxx = fyy = fxy = fyx = 0. 6), d2 z = 0. 9) is satisfied.

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