04Final Exam Solutions

# 04Final Exam Solutions - MATH150 Final Exam Solutions 1(12...

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Unformatted text preview: MATH150 Final Exam Solutions 12/9/04 1. (12 marks) Evaluate the following limits or show that they don’t exist: (a) lim x → 1- sec x tan 2 x (b) lim x →∞ (ln x ) 1 /x (c) lim ( x,y ) → (2 , 1) x 2- 4 y 2 x- 2 y Solution: (a) (4 marks) lim x → 1- sec x tan 2 x ” ” = lim x →- sec x tan x 2 tan x sec 2 x = lim x →- 1 2 sec x =- 1 / 2 (b) (5 marks) lim x →∞ ln ‡ (ln x ) 1 /x · = lim x →∞ ln ln x x ” ∞ ∞ ” = lim x →∞ 1 ln x · 1 x 1 = lim x →∞ 1 x ln x = 0 Thus lim x →∞ (ln x ) 1 /x = lim x →∞ e ln ( (ln x ) 1 /x ) = e lim x →∞ ln ( (ln x ) 1 /x ) = e = 1 (c) (3 marks) lim ( x,y ) → (2 , 1) x 2- 4 y 2 x- 2 y = lim ( x,y ) → (2 , 1) ( x- 2 y )( x + 2 y ) x- 2 y = lim ( x,y ) → (2 , 1) x + 2 y = 4 2. (10 marks) (a) Apply the mean value theorem to the function f ( x ) = √ x on the interval [49 , 50] to show that √ 50 = 7 + 1 2 √ c for some number c in (49 , 50). (b) Use (a) to show that 7 < √ 50 < 99 14 . Solution: (a) (6 marks) f ( x ) = √ x is differentiable. If we apply the MVT to [ a, b ] = [49 , 50] we get f ( b )- f ( a ) b- a = √ 50- √ 49 50- 49 = √ 50- 7 = f ( c ) = 1 2 √ c for some c in (49 , 50). Thus √ 50 = 7 + 1 2 √ c for some c in (49 , 50). (b) (4 marks) By (a) we have √ 50 = 7 + 1 2 √ c for some c in (49 , 50). Since the square root function is monotonically increasing we get 7 < √ 50 < 7 + 1 2 √ 49 = 7 + 1 14 = 99 14 3. (12 marks) Let f ( x ) = x 2 / 3 (5- x ), defined for all real numbers x . (a) Find and classify the critical and singular points (if any) and find the intervals of increase and decrease of f . (b) Find all inflection points (if any) and the intervals of concavity of f . (c) Find the global maximum and minimum values of f on the interval [- 1 , 4]. Solution: (a) (4 marks) f ( x ) = x 2 / 3 (5- x ) = 5 x 2 / 3- x 5 / 3 . Thus f ( x ) = 10 3 x- 1 / 3- 5 3 x 2 / 3 = 5 3 x- 1 / 3 (2- x ) Thus x = 0 is a singular point and x = 2 is a critical point. Furthermore f ( x ) < for x < > for 0 < x < 2 < for x > 2 Hence x = 0 is a local minimum, x = 2 is a local maximum, f is decreasing on (-∞ , 0) and (2 , ∞ ) and is increasing on (0 , 2). (b) (5 marks) f ( x ) =- 10 3 x- 4 / 3- 10 3 x- 1 / 3 =- 10 3 x- 4 / 3 (1 + x ). Candidates for inflection points are thus x =- 1 and x = 0. Since x- 4 / 3 > 0 for all x = 0, we have f ( x ) > for x <- 1 < for- 1 < x < < for x > Thus x =- 1 is an inflection point whereas...
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## This note was uploaded on 08/23/2008 for the course MATH 150 taught by Professor Hundemer during the Fall '04 term at McGill.

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04Final Exam Solutions - MATH150 Final Exam Solutions 1(12...

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