# practice test 3 answers - MAT 111 Practice Test 3 Solutions...

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Unformatted text preview: MAT 111 Practice Test 3 Solutions 1 1. (25 points) Find the absolute maximum and minimum values of f on the given interval. (a) (15 points) f ( x ) = x 3- 1 . 5 x 2- 6 x + 1 , [- 2 , 4] solution: f ( x ) = 3 x 2- 3 x- 6 = 3( x 2- x- 2) = 3( x- 2)( x + 1) So, critical points are when f ( x ) = 0. That is x = 2 and x =- 1. Evaluating f at critical points and end points: f (- 2) = (- 2) 3- 1 . 5(- 2) 2- 6(- 2) + 1 =- 8- 6 + 12 + 1 = 1 f (- 1) =- 1- 1 . 5 + 6 + 1 = 4 . 5 f (2) = 8- 6- 12 + 1 =- 9 f (4) = 64- 24- 24 + 1 = 17 So, absolute minimum is- 9 at x = 2 Absolute Maximum is 17 at x = 4. (b) (10 points) f ( x ) = x + 2 sin x, [- π, π ] Solution: f ( x ) = 1 + 2 cos x , so critical points are when 1 + cos x = 0 that is when cos x =- 1 / 2. In the interval [- π, π ], this happens for x = 2 π/ 3 and x =- 2 π/ 3. Evaluating f at critical points and end points: f (- π ) = 1 + 2 sin- π = 1 + 2(0) = 1 f (- 2 π/ 3) = 1 + 2 sin (- 2 π/ 3) = 1 + 2(- √ 3 / 2) = 1- √ 3 f (2 π/ 3) = 1 + 2 sin (2 π/ 3) = 1 + 2( √ 3 / 2) = 1 + √ 3 f ( π ) = 1 + 2 sin π = 1 + 2(0) = 1...
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practice test 3 answers - MAT 111 Practice Test 3 Solutions...

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