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M408C - rev1sol

# M408C - rev1sol - M408C Calculus 5 Fall 2002 Solutions of...

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M408C Calculus 55815/20/25 Fall 2002 Solutions of Midterm 1 Review Problems 1. (a) lim x 1 x 2 - 2 x + 1 x 2 - 1 = lim x 1 ( x - 1) 2 ( x - 1)( x + 1) = lim x 1 x - 1 x + 1 = 0 2 = 0 (b) lim x 2 + f ( x ) = lim x 2 + ( x 2 - x ) = 2 2. (a) lim x 0 7 x 2 sin 2 5 x = lim x 0 (5 x ) 2 sin 2 5 x · 7 25 = 7 25 (b) lim x 0 tan 3 x x 2 + 4 x = lim x 0 tan 3 x 3 x · 3 x 4 x + x 2 = 1 · lim x 0 3 4 + x = 3 4 3. (a) We have f (1 + ) = lim x 1 + (2 x 3 + 1) = lim x 1 + 6 x 2 = 6 f (1 - ) = lim x 1 - (3 x 2 ) = lim x 1 - 6 x = 6 Since f (1 - ) = f (1 + ), f is differentiable at 1 and f (1) = 6. (b) No, f ( x ) = | 2 x - 5 | is not differentiable at x = 5 / 2. 4. (a) Using the product and chain rules, ( ( x 2 + 1) 2 ( x - 1) 3 ) = 2( x 2 + 1)2 x · ( x - 1) 3 + ( x 2 + 1) 2 · 3( x - 1) 2 = ( x 2 + 1)( x - 1) 2 (4 x ( x - 1) + 3( x 2 + 1)) = ( x 2 + 1)( x - 1) 3 (7 x 2 - 4 x + 3) (b) Using the quotient rule, dy dx = d dx x 2 - 3 x + 1 x + 2 = (2 x - 3)( x + 2) - ( x 2 - 3 x + 1) · 1 ( x + 2) 2 = x 2 + 4 x - 7 ( x + 2) Thus, at x = 2 we have dy dx = 5 16 . 5. We have dy du = 1 · (1 - u ) - (1 + u ) · ( - 1) (1 - u ) 2 = 2 (1 - u ) 2 , du dt = 2 3 t - 1 / 3 .

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M408C Calculus 55815/20/25 Fall 2002 Thus, by the chain rule, dy dt = dy du · du dt = 2 (1 - u ) 2 · 2 3 t - 1 / 3 = 4 3 t 1 / 3 (1 - t 2 / 3 ) 2 6. y = ( x sin(2 x + 1)) = sin(2 x + 1) + 2 x cos(2 x + 1) y = (sin(2 x + 1) + 2 x cos(2 x + 1)) = 2 cos(2 x + 1) + 2 cos(2 x + 1) - 4 x sin(2 x + 1) = 4(cos(2 x + 1) - x sin(2 x + 1)) 7. At x = π/ 4 we have y = cos 2 ( π/ 4) = 1 / 2 and dy dx = - 2 cos x sin x = - 2 1 2 1 2 = - 1 . Thus the equation of the tangent line is y - 1 2 = - x - π 4 . 8. Differentaiting implicitly 2 x 2 + 2 xy + y 2 = 2 we obtain d dx (2 x 2 + 2 xy + y 2 ) = 0 4 x + 2 y + 2 x dy dx + 2
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