sample-test1-fall-2005

sample-test1-fall-2005 - Math 221 Test 1 1(9 points Given...

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Math 221 Test 1 September 21, 2005 1. (9 points) Given f ( x ) = sin x and g ( x ) = x + 1, fnd Formulas For the Following Functions, and fnd their domains: (a) ± f g ² ( x ) = Solution: sin x x + 1 The domain oF f g is Solution: ( - 1 , ) or { x : x > - 1 } (b) ( f g )( x ) Solution: sin( x + 1) The domain oF f g is Solution: [ - 1 , ) or { x : x ≥ - 1 } (c) ( g f )( x ) = Solution: sin x + 1 The domain oF g f is Solution: R or ( -∞ , ) or “the set of all real numbers” 2. (8 points) Suppose f , g and h all have the same domain and lim x a f ( x ) = 0, lim x a g ( x ) = - 1, and lim x a h ( x ) = 10. ±ind the Following limits. IF the limit does not exist then mark it as “DNE” or or -∞ , whichever is the most inFormative. (a) lim x a [ f ( x ) - g ( x )] = Solution: lim x a [ f ( x ) - g ( x )] = lim x a f ( x ) - lim x a g ( x ) = 0 - ( - 1) = 1 (b) lim x a [ - g ( x ) · h ( x )] = Solution: lim x a [ - g ( x ) · h ( x )] = - lim x a g ( x ) · lim x a h ( x ) = - ( - 1) · 10 = 10 (c) lim x a p g ( x ) + h ( x ) = Solution: lim x a p g ( x ) + h ( x ) = q lim x a g ( x ) + lim x a h ( x ) = - 1 + 10 = 9 = 3 (d) lim x a ³ g ( x ) ( f ( x )) 2 ´ = 1
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2 Solution: ± g ( x ) ( f ( x )) 2 ² = -∞ since the limit of the numerator is negative and the limit of the denominator is 0 and the denominator is positive, since it is the square of f ( x ). [Technically, there are some cases in which the limit does not exist.] 3. (20 points) Let f ( x ) = - 8 if x < - 2 x 3 if - 2 x < - 1 x + 1 if - 1 x < 1 4 if x = 1 3 - x 2 if 1 < x (a) Does lim x →- 2 f ( x ) exist? If so, what is it? In either case, justify your answer. Solution: lim x →- 2 f ( x ) exists and equals - 8, since the left- and right-hand limits exist and both equal - 8: lim x →- 2 - f ( x ) = lim x →- 2 - ( - 8) = - 8 and lim x →- 2 + f ( x ) = lim x →- 2 + x 3 = ( - 2) 3 = - 8 (b) Does lim x →- 1 f ( x ) exist? If so, what is it? In either case, justify your answer. Solution: lim x →- 1 f ( x ) does not exist since the left- and right-hand limits exist but are not equal: lim x →- 1 - f ( x ) = lim x →- 1 - x 3 = ( - 1) 3 = - 1 but lim x →- 1 + f ( x ) = lim x →- 1 + ( x + 1) = ( - 1) + 1 = 0 (c) Does lim x 1 f ( x ) exist? If so, what is it? In either case, justify your answer. Solution: lim x 1 f ( x ) exists and equals 2, since the left- and right-hand limits exist and both equal 2: lim x 1 - f ( x ) = lim x 1 - ( x + 1) = 1 + 1 = 2 and lim x 1 + f ( x ) = lim x 1 + (3 - x 2 ) = 3 - 1 2 = 2 (d) List all discontinuities of f ; that is, the x -values where f is not continuous:
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This note was uploaded on 03/13/2009 for the course MATH 1214-008 taught by Professor Maier during the Fall '08 term at UNC Charlotte.

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sample-test1-fall-2005 - Math 221 Test 1 1(9 points Given...

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