1007_Tut3_Sol

1007_Tut3_Sol - MATH 1007 A Tutorial #3 Oct 27, 2006...

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Unformatted text preview: MATH 1007 A Tutorial #3 Oct 27, 2006 SOLUTIONS You must show your work for all questions! 1. [3 marks] Find the most general antiderivative: (a) f ( x ) = 4 + x 2- 5 x 3 Solution f ( x ) = 4 x + 1 3 x 3- 5 4 x 4 + C (b) g ( x ) = x 20 + 4 x 10 + 8 Solution G ( x ) = 1 21 x 21 + 4 11 x 11 + 8 x + C 2. [7 marks] Let y = f ( x ) = x (ln x ) 2 . Note that the domain of f is (0 , ∞ ) since ln x is only defined for x > 0. The following approximate values may be useful e ≈ 2 . 718 e 2 ≈ 7 . 389 1 e ≈ . 368 1 e 2 ≈ . 135 4 e 2 ≈ . 541 Please answer ALL of the following questions: (separate sheet provided) (a) Find the x intercept(s) (set y = 0). [Note: there is no y intercept.] Solution [0.5 marks] y = 0 = ⇒ x (ln x ) 2 = 0 = ⇒ x = 0 which is not in domain, or (ln x ) 2 = 0 = ⇒ ln x = 0 = ⇒ x = 1 (b) Evaluate lim x →∞ f ( x ) to see what happens when x is large. Solution [0.5 marks] lim x →∞ f ( x ) = lim x →∞ x (ln x ) 2 → ∞ (Both x and ln x approach ∞ as x → ∞ .) (c)...
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This note was uploaded on 03/22/2010 for the course MATH 1007 taught by Professor Unknown during the Fall '07 term at Carleton.

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1007_Tut3_Sol - MATH 1007 A Tutorial #3 Oct 27, 2006...

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