131f08x1ANS

131f08x1ANS - Math 131 Exam 1 Solutions 1(a[8 avg v = h(5 2...

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Unformatted text preview: Math 131 Exam 1 Solutions 10/1/08 1. (a) [8%] avg v = h (5 . 2) − h (5) 5 . 2 − 5 = [1200 − 4 . 9(5 . 2) 2 ] − [1200 − 4 . 9(5) 2 ] . 2 = − 4 . 9[(5 . 2) 2 + (5) 2 ] . 2 = − 4 . 9 2 . 04 . 2 = − 49 . 98 m/s . (b) [8%] avg v = h (5 + Δ t ) − h (5) Δ t = [1200 − 4 . 9(5 + Δ t ) 2 ] − [1200 − 4 . 9(5) 2 ] Δ t (m/sec) The question did not ask you to simplify the result, but if you do so, you should obtain: = − 4 . 9 ( 25 + 10Δ t + (Δ t ) 2 ) + 4 . 9(25) Δ t = − 4 . 9(10Δ t + (Δ t ) 2 Δ t = − 4 . 9Δ t (10 + Δ t ) Δ t = − 4 . 9(10 + Δ t ) (m/sec) 2. (a) [10%] The desired slope m is given by: m = lim h → f (4 + h ) − f (4) h = lim h → [(4 + h ) 2 − 3(4 + h )] − [4 2 − 3(4)] h = lim h → 16 + 8 h + h 2 − 12 − 3 h − 16 + 12 h = lim h → 5 h + h 2 h = lim h → h (5 + h ) h = lim h → (5 + h ) = 5 + 0 = 5 . (b) [6%] The form of the tangent line is y − f (4) = m ( x − 4) . The tangent line has slope m = 5 and goes through the point ( 4 , f (4) ) = ( 4 , 4 2 − 3(4) ) = (4 , 4) . So its equation is y − 4 = 5( x − 4) , That form is OK. So is a simplified form such as: y = 5 x − 16 1 3. (a) [8%] Factor the denominator to obtain f ( x ) = 4 x 2 + 5 ( x − 1) ( x − 5) ....
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131f08x1ANS - Math 131 Exam 1 Solutions 1(a[8 avg v = h(5 2...

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