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MATH 140 Dec 2006 Answers

MATH 140 Dec 2006 Answers - y(300-y 2 subject to< y<...

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Q1 (a) 0 , (b) 2 , (c) e - 2 , (d) NO FINITE OR INFINITE LIMIT, (e) - 3 2 . Q2 (a) y = 3 and y = - 1 , (b) NONE, (c) Increasing at a rate of 1 12 radi- ans/second, (d) x = 0 and x = 3 (note: this topic not covered in Fall 2010). Q3 (a) - 1 2 x - 3 2 - x - 4 3 , (b) u - 1 , (c) arctan( x ) + 3 , (d) arcsin( t ) . Q4 (a) lim x a f ( x ) - f ( a ) x - a , (b) f is not differentiable at x = 1 since lim x 1 | 1 - x | - 0 x - 1 does not exist, since lim x 1 - | 1 - x | - 0 x - 1 = - 1 6 = 1 = lim x 1+ | 1 - x | - 0 x - 1 . (c) - 1 . Q5 b = 14 3 , c = 0 , a = - 2 3 . Q6 We need to show that 0 = 2 y - 4 + e y has exactly one solution in y . Let f ( y ) = 2 y - 4+ e y , then f (0) = - 3 and f (1) = e - 2 > 0 . Since f is continuous we see from the Intermediate Value Theorem that there exists y with 0 < y < 1 such that f ( y ) = 0 . On the other hand, if f ( y 1 ) = f ( y 2 ) with y 1 < y 2 , then according to the Mean Value Theorem applied to the everywhere differentiable function f we have 0 = f ( y 2 ) - f ( y 1 ) y 2 - y 1 = f 0 ( η ) = 2 + e η > 0 , for some η with y 1 < η < y 2 . This (i.e. 0 > 0 ) is a contradiction. Hence there is only one distinct solution. Q7 First decoding: maximize xy subject to x + y 2 = 300 , x > 0 , y > 0 . We decide to use the constraint to solve for
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Unformatted text preview: y (300-y 2 ) subject to < y < √ 300 . If f = y (300-y 2 ) = 300 y-y 3 , then df dy = 300-3 y 2 . The maximum occurs when y = 10 and consequently x = 200 . Q8 (a) increasing everywhere, (b) One critical point at x = 1 , f (1) = 1-ln(2) , type other. No local extrema. (c) x = 1 is a point of inflection, i.e. (1 , 1-ln(2)) in the xy-plane. x =-1 is another point of inflection i.e. (-1 ,-1-ln(2)) in the xy-plane. (d) Sketch and details below. > f:=x - ln(1+xˆ2); f := x-ln(1 + x 2 ) > g:=diff(f,x); g := 1-2 x 1 + x 2 1 > g:=factor(g); g := (-1 + x ) 2 1 + x 2 > solve(g,x); 1 , 1 > h:=diff(g,x); h :=-2 1 + x 2 + 4 x 2 (1 + x 2 ) 2 > h:=factor(h); h := 2(-1 + x )(1 + x ) (1 + x 2 ) 2 > solve(h,x); 1 ,-1 > plot(f,x=-2. .4); –3 –2 –1 1 –2 –1 1 2 3 4 x 2...
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