# 2001test1 - (a[5 points f x = x x f x(b[5 points x 5-xy y 3 = 8 y(0(c[5 points f x =(1 x 2 3(2-x 1 3(1 x 2 3 2(1 ln x 1 2 f(1(d[5 points h x =

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University of Toronto at Scarborough Physical Sciences Division, Mathematics First Midterm Test MATA26Y Calculus Examiners: X. Jiang R. Sharpe S. Tryphonas Date: November 14, 2001 Duration: 110 minutes 1. Evaluate the limit: (a) [5 points] lim x 1 2 x 2 + 2 x - 4 x - 1 (b) [5 points] lim x 0 1 - cos2 x 12 x 2 (c) [5 points] lim x →∞ sin x x 2. Let f ( x ) = x 2 + 2 x x 3 + 4 x 2 (a) [3 points] Find lim x 0 + f ( x ) (b) [2 points] Find lim x 0 - f ( x ) 3. [10 points] For f ( x ) = 2 x 3 x 2 - 1 , ﬁnd positive number R, m,M such that m | x | r ≤ | f ( x ) | ≤ M | x | r for all x satisfying | x | ≥ R where r = ord f . 4. [10 points] Find all possible values of a and b so that f ( x ) is continuous for all x R if f ( x ) = | ax + 3 | if x ≤ - 1 | 3 x + a | if - 1 < x 0 b sin2 x x - 2 b if 0 < x < π cos 2 x - 3 if x π

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MATA26Y page 2 5. (a) [4 points] Give the statement of the Mean Value Theorem. (b) [6 points] If h 0 ( x ) = 1 3 + 2 x 2 for all x R and h (1) = 0, show 1 11 < h (2) < 1 5 . 6. Find the indicated derivatives
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Unformatted text preview: (a) [5 points] f ( x ) = x x ; f ( x ) (b) [5 points] x 5-xy + y 3 = 8 ; y (0) (c) [5 points] f ( x ) = (1 + x ) 2 / 3 (2-x ) 1 / 3 (1 + x 2 ) 3 / 2 (1 + ln x ) 1 / 2 ; f (1) (d) [5 points] h ( x ) = tan(arccos x ) ; h ( x ) 7. Let f ( x ) = x 2 + 2 e 2( x-1) . (a) [5 points] Find the linear approximation A 1 ( x ) for f ( x ) at x = 1. (b) [10 points] Find h > 0 such that | f ( x )-A 1 ( x ) | < . 01 for | x-1 | < h . 8. Let P ( x ) = 10 x 4-5 x-4 (a) [5 points] How many roots does P ( x ) have? (b) [10 points] Determine the smallest root of P ( x ) to four decimal place accuracy (i.e., to within an error of ± . 00005)...
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## This note was uploaded on 10/04/2011 for the course MATH 16121 taught by Professor Rachelbelinsky during the Spring '11 term at Georgia State University, Atlanta.

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2001test1 - (a[5 points f x = x x f x(b[5 points x 5-xy y 3 = 8 y(0(c[5 points f x =(1 x 2 3(2-x 1 3(1 x 2 3 2(1 ln x 1 2 f(1(d[5 points h x =

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