M408C - Review 2 SOlutions

# M408C - Review 2 SOlutions - M408C Calculus 5 Fall 2002...

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Unformatted text preview: M408C Calculus 55815/20/25 Fall 2002 Midterm 2 Review Problems and Solutions The exam will cover the following sections of Salas and Hille’s Calculus , 8th edition: 4.3–4.8, 5.1–5.7, 6.1 1. Find the critical numbers and classify the extreme values (a) f ( x ) = x p 4- x 2 (b) f ( x ) = tan x- 2 x,- π 3 ≤ x < π 2 Solution. (a) First of all, the domain of f is the interval [- 2 , 2]. Next, f ( x ) = p 4- x 2 + x ·- x √ 4- x 2 = 4- x 2- x 2 √ 4- x 2 = 2(2- x 2 ) √ 4- x 2 for x ∈ (- 2 , 2). Hence the critical points are solutions of f ( x ) = 0 ⇔ x = ± √ 2. Using the method of intervals we find also that f < 0 in (- 2 ,- √ 2) , f > 0 in (- √ 2 , √ 2) , f < 0 in ( √ 2 , 2) . Thus x =- 2 endpoint max f (- 2) = 0 x =- √ 2 local min f (- √ 2) =- 2 x = √ 2 local max f ( √ 2) = 2 x = 2 endpoint min f (2) = 0 Also, we see that x =- √ 2 is absolute minimum and x = √ 2 is absolute maximum. (b) Find the derivative f ( x ) = sec 2 x- 2 , x ∈ (- π/ 3 ,π/ 2) Thus the critical points are solutions of f ( x ) = 0 ⇔ sec x = ± √ 2 ⇔ x = ± π/ 4....
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M408C - Review 2 SOlutions - M408C Calculus 5 Fall 2002...

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