V63_0121_0001_2008F_Midterm_I_Solutions

V63_0121_0001_2008F_Midterm_I_Solutions - x (d) y = (6...

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SOLUTIONS TO MIDTERM I V63.0121.001.2008F 1. (a) y = 7 x - 5. (b) ( 3 2 , 19 4 ), y = 7 x - 23 4 . 2. (a) Graph. (b) This function is continuous on ( -∞ , + ) and is differen- tiable on ( -∞ , 1) (1 , + ) because lim h 0 + f (1+ h ) - f (1) h = 1 but lim h 0 - f (1+ h ) - f (1) h = - 1. (c) a = - 2. (d) F is not differentiable at x = 1. Since lim h 0 + F (1+ h ) - F (1) h = 2 but lim h 0 - F (1+ h ) - F (1) h = - 1. 3. (a) 5 (b) -7 (c) 1. 4. (a) y 0 = 15 x 4 - 15 2 x 3 2 - x - 3 2 (b) y 0 = 2 cos x - csc 2 x (c) y 0 = (2 x + 3) sin x + ( x 2 + 3 x + 1) cos
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Unformatted text preview: x (d) y = (6 x-2)( x 2 +4)-2 x (3 x 2-2 x +1) ( x 2 +4) 2 = 2 x 2 +22 x-8 ( x 2 +4) 2 . 5. For x >-1, lim h f ( x + h )-f ( x ) h = lim h x +1+ h- x +1 h = lim h x +1+ h-x-1 h ( x +1+ h + x +1) = lim h h h ( x +1+ h + x +1) = 1 2 x +1 . 1...
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This note was uploaded on 10/25/2010 for the course MATH V63.-0121- taught by Professor Staff during the Fall '08 term at NYU.

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