Part (a) dy/dx = 5(-1)2 - = = d2y/dx2 =

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Presentation transcript:

Part (a) 6 -9 6 dy/dx = 5(-1)2 - = 5 - -1 = -4 - 2 d2y/dx2 = = 5 - -1 = 6 d2y/dx2 = 10x + 6 (y-2)-2 (dy/dx) = 10 (-1) + 6 (-4-2)-2 (6) -9 = -10 + 1 =

Therefore, f(x) is never tangent to the x-axis. Part (b) For this to occur, dy/dx would need to be zero at the same point where y = 0. For dy/dx to be zero, 5x2 = 6 y-2 For both dy/dx and y to be zero, 5x2 = -3 Since 5x2 can’t be negative, this is impossible. Therefore, f(x) is never tangent to the x-axis.

From Part (a), f’(-1) = 6 and f”(-1) = -9 Part (c) From Part (a), f’(-1) = 6 and f”(-1) = -9 P2(x) = f(-1) + f’(-1) (x+1)1 1 ! f”(-1) (x+1)2 2 ! + P2(x) = 6 (x+1) -9 (x+1)2 2 + -4 +

Part (d) (0,5/8) 5 (-1)2 - 6 -4 - 2 5 (-0.5)2 - 6 -1 - 2 (-1,-4) 6 0.5 (x,y) dy/dx Dx Dy = (dy/dx)(Dx) (x+Dx,y+Dy) (-1,-4) 6 0.5 3 (-0.5,-1) (-0.5,-1) 13/4 0.5 13/8 (0,5/8)