Midterm questions

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Department
Mathematics
Course
MATA30H3
Professor
Sophie Chrysostomou
Semester
Fall

Description
University of Toronto Scarborough Department of Computer & Mathematical Sciences MATA30: Calculus I - Solution Midterm Test 1. Let f(x)=2 e ▯x +3. (a) [2m rs] Find the domain and range of f. solution : Since e ▯x is defined for all x ▯ R, then the domain of f(x)=2 e ▯x +3 is R. x x ▯x ▯x 0 1 |1 ▯ x| 1 ▯ x ▯(x ▯ 1) ▯1 lim = lim = lim = lim = ▯1 x▯1▯ x ▯ x x▯1▯ x(x ▯ 1) x▯1▯ x(x ▯ 1) x▯1▯ x |1 ▯ x| x ▯ 1 1 lim = lim = lim =1 x▯1+ x ▯ x x▯1+ x(x ▯ 1) x▯1+x |1 ▯ x| |1 ▯ x| |1 ▯ x| x▯1▯ x ▯ x ▯=x▯1+ x ▯ x thereforex▯1mx ▯ x does not exist. ▯ ▯ 2 x+1 (d) x▯▯1 (x +1) arctan e ▯ ▯ Since ▯ < arctanx< for all xe: 2 2 ▯ ▯ ▯ x+1 ▯ ▯ 2 < arctan e < 2 for all x ▯= ▯1( ▯) 2 2 Since (x +1) > 0 for all x ▯= ▯1, then, multiplying all sides of (▯) with (x +1) gives: ▯(x +1) 2 ▯ 1 ▯ ▯(x +1) 2 ▯ < (x +1) arctan e x+1 < for all x ▯= ▯1 2 2 2 2 ▯(x +1) ▯(x +1) Since lim ▯ = 0 and lim =0,therytheSqueeze x▯▯1 2 x▯▯1 2 theorem: ▯ 1 ▯ lim (x +1) arctan ex+1 =0 x▯▯1 ——————————————————————————— Alternatively (A) 2 lim (x +1) =0 x▯▯1 And ▯ 1 ▯ lim arctan ex+1 =0 x▯▯1▯ 1 x+1 (x +1 < 1,x +1 ▯ 0, so ▯▯▯ and e ▯ 0) x +1 Thus ▯ 1 ▯ lim (x +1) arctan ex+1 =0 x▯▯1 ▯ And (B) A ▯ 1 ▯ ▯ lim arctan e x+1 = x▯▯1+ 2 1 1 (x +1 > 1,x +1 ▯ 0, so ▯▯ and e x+1▯▯ ) x +1 Thus ▯ ▯ 2 x+1 ▯ x▯▯1+(x +1) arctan e =0 · 2 =0 From (A) and (B), Thus ▯ ▯ ▯ ▯ lim (x +1) arctan ex+1 = lim (x +1) arctan ex+1 =0 x▯▯1▯ x▯▯1+ 4 ▯ 2 x ▯ 1+1 4. [10 marks]Lt f(x)= x ▯ 1 . (a) [2maks ] Give the doma▯n of f. 2 solution : x ▯ dom(f) is x ▯ 1 is defined and if x ▯ 1 ▯=0. So x ▯=1and x ▯ 1 ▯ 0
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