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1. Calculate each it below, if it eists. If a it does not eist, eplain why. Show all work. Writing an answer with no justification may not yield any marks. 2 2 1 [3] (a) 1 1. [5] (b) 3 Solution: 22 1 = 1 and 1 = ( 1) = 0+. 1 1 1 Therefore the it goes to either ±. Since the fraction is positive, f() =. 1 + 3 2. 3 Solution: We rationalize the numerator: 3 + 3 2 3 ( + 3 2)( + 3 + 2) = 3 ( 3)( + 3 + 2) = 3 + 3 2 ( 3)( + 3 + 2) = 3 ( 3) ( 3)( + 3 + 2) 1 = 3 + 3 + 2 = 1 2 6. [6] (c) 2 + 3 3 + 1. Solution: Notice that 2 = because < 0. 2 (1 + 3) 2 + 3 3 + 1 = = = (1 + 0) (3 + 0) = 3 + 1 (1 + 3) (3 + 1) = = 1 3. 2 (1 + 3 ) (3 + 1) (1 + 3) (3 + 1 )
2 2 [2] (d). 2 2 4 Solution: ( [4] (e) ( 2) 4 sin 2 2 2 2 2 4 3 ( 2) 2 ). ( 2)( + 1) = 2 ( + 2)( 2) = ( + 1) 2 ( + 2) = 3 4 ( ) 3 Solution: For all 2, 1 sin 1. ( 2) 2 Thus, for 2, ( ) ( 2) 4 ( 2) 4 3 sin ( 2) 4. ( 2) 2 Since both ( 2) 4 0 and ( 2) 4 0, as 2, then by the Squeeze Theorem, ( ) ( 3 2 2)4 sin = 0. ( 2) 2
[7] 2. Let f be the function: (3k 2) if < 1 f() = 1 if = 1. k 2 if > 1 Find all values of k for which f() is continuous for all real numbers. Be sure to fully justify your answer. Solution: For any k, the function is continuous on (, 1) (1, ) because it is a polynomial in those intervals. The function will be continuous at = 1 if Furthermore, 1 1 1 f() = (3k 2) 1 f() = f() = f(1) = 1. + f() = k2 1 + 1 + = 3k 2. = k 2. Then, We now find k: 3k 2 = 1 = k 2. 3k 2 = 1 k 2 = 1 3k = 3 k = ± 1 k = 1 k = ±1. The only value of k that makes both above statements true is k = 1. We conclude that for k = 1 the function f() is continuous for all real numbers.
[7] 3. Use the definition of derivative to find f (). No credit will be given for any other method. f() = 1 3 Solution: f f( + h) f() () = h 0 h = h 0 1 3 ( + h) 1 3 h 3 (3 ( + h)) = h 0 h(3 ( + h))(3 ) = h 0 = h 0 3 3 + + h h(3 h)(3 ) h h(3 h)(3 ) = h 0 1 = (3 h)(3 ) 1 (3 ) 2.
[5] 4. Find the domain of the function f() = of intervals. 5(9 2 ). Epress your final answer in terms ( 9) 5(9 2 ) 5(3 )(3 + ) Solution: f() = =. ( 9) ( 9) The radicand needs to be non-negative under the square root and non-zero in the 5(3 )(3 + ) denominator. Let h() = ( 9) < 3 3 < < 3 3 < < 9 > 9 3 + + 3 + + + + 9 + h() + + Therefore, D f = (, 3] [3, 9). [5] 5. Use the Intermediate Value Theorem to show that 3 1 + 2 solution on the interval [0, 2]. = 0 has at least one Solution: Let f() = 3 1, then f() is continuous for all ecept = 2. + 2 In particular, f() is continuous on the interval [0, 2]. We now check the value of f() at the end points: f(0) = 0 3 0 1 0 + 2 = 1 2 < 0, f(2) = 23 2 1 2 + 2 = 23 4 > 0 Since the values of the function at the end points have different signs, using the Intermediate Value Theorem, we conclude that there is a point c on the interval [0, 2] such that f(c) = 0. Therefore, there eist at least one solution of the equation on [0, 2].
[6] 6. Use the definition of derivative to show that { 2 + 2 if 1 f() = + 2 if > 1 is not differentiable at = 1. Solution: The right-hand it of the difference quotient of f at = 1 is given by f(1 + h) f(1) h 0 + h = h 0 + [(1 + h) + 2] 3 h = h 0 + 1 = 1. The left-hand it of the difference quotient of f at = 1 is given by f(1 + h) f(1) h 0 h = h 0 [(1 + h) 2 + 2] 3 h = h 0 (2 + h) = 2. Since the left-hand and right-hand its of the difference quotient differ at = 1, the function f is not differentiable there.
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[Scrap page] If you are using this page to continue your work from a previous question, clearly indicate on the original page that your work is continuing here. Otherwise, your work will not be marked.