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# Consider the equation below- f(x) - 6 cos2x - 12 sin x- 0 - x - 2-.docx

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# Consider the equation below- f(x) - 6 cos2x - 12 sin x- 0 - x - 2-.docx

Consider the equation below. f(x) = 6 cos2x ? 12 sin x, 0 ? x ? 2? find : the interval on which f is increasing the interval on which f is decreasing the local minimum and maximum values of f. the inflection points. the interval on which f is concave up the interval on which f is concave down
Solution
f\'(x) = - 14 sin(x) cos(x) - 14 cos(x) = -14 cos(x) ( sin(x) + 1) So f\'(x)=0 when : cos(x) = 0 => x = pi/2 & 3pi/2 & sin(x) = -1 => x = 3pi/2 when x < pi/2 f\'(x) is negative when pi/2 < x < 3pi/2 f\'(x) is positive when 3pi/2 < x < 2pi f\'(x) is negative So : a) the interval on which f is increasing ( pi/2 , 3pi/2) b) the intervals on which f is decreasing [0 , pi/2) & (3pi/2 , 2pi] c) f\'\'(x) = -14 [ cos^2(x) - sin(x) (sin(x)+1)] at x = pi/2 f\'\'(x) is positive so there is a local minimum at x = pi/2 the local minimum = -14 at x = 3pi/2 f\'\'(x) =0 so we study the sign of f\'(x) around this value it changes from +ve to -ve so there is a local maximum at x = 3pi/2 the local maximum = 14 d) putting f\'\'(x) = 0 [ cos^2(x) - sin(x) (sin(x)+1)] = 0 1- sin^2(x) - sin^2(x) - sin(x)=0 2 sin^(x) + sin (x) -1 = 0 (2 sin(x) -1)(sin(x) +1) = 0 sin(x) = 0.5 => x = pi/6 & 5pi/6 sin(x) = -1 => x = 3pi/2 testing the sign of f\'\'(x) around these points we get : there are two inflection points at x = pi/6 , 5pi/6 e) the curve is concave up in the interval (0 , pi/6 ) & ( 5pi/6 , 2pi) the curve is concave down in the interval (pi/6 , 5pi/6 )
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Consider the equation below. f(x) = 6 cos2x ? 12 sin x, 0 ? x ? 2? find : the interval on which f is increasing the interval on which f is decreasing the local minimum and maximum values of f. the inflection points. the interval on which f is concave up the interval on which f is concave down
Solution
f\'(x) = - 14 sin(x) cos(x) - 14 cos(x) = -14 cos(x) ( sin(x) + 1) So f\'(x)=0 when : cos(x) = 0 => x = pi/2 & 3pi/2 & sin(x) = -1 => x = 3pi/2 when x < pi/2 f\'(x) is negative when pi/2 < x < 3pi/2 f\'(x) is positive when 3pi/2 < x < 2pi f\'(x) is negative So : a) the interval on which f is increasing ( pi/2 , 3pi/2) b) the intervals on which f is decreasing [0 , pi/2) & (3pi/2 , 2pi] c) f\'\'(x) = -14 [ cos^2(x) - sin(x) (sin(x)+1)] at x = pi/2 f\'\'(x) is positive so there is a local minimum at x = pi/2 the local minimum = -14 at x = 3pi/2 f\'\'(x) =0 so we study the sign of f\'(x) around this value it changes from +ve to -ve so there is a local maximum at x = 3pi/2 the local maximum = 14 d) putting f\'\'(x) = 0 [ cos^2(x) - sin(x) (sin(x)+1)] = 0 1- sin^2(x) - sin^2(x) - sin(x)=0 2 sin^(x) + sin (x) -1 = 0 (2 sin(x) -1)(sin(x) +1) = 0 sin(x) = 0.5 => x = pi/6 & 5pi/6 sin(x) = -1 => x = 3pi/2 testing the sign of f\'\'(x) around these points we get : there are two inflection points at x = pi/6 , 5pi/6 e) the curve is concave up in the interval (0 , pi/6 ) & ( 5pi/6 , 2pi) the curve is concave down in the interval (pi/6 , 5pi/6 )
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### Consider the equation below- f(x) - 6 cos2x - 12 sin x- 0 - x - 2-.docx

1. 1. Consider the equation below. f(x) = 6 cos2x ? 12 sin x, 0 ? x ? 2? find : the interval on which f is increasing the interval on which f is decreasing the local minimum and maximum values of f. the inflection points. the interval on which f is concave up the interval on which f is concave down Solution f'(x) = - 14 sin(x) cos(x) - 14 cos(x) = -14 cos(x) ( sin(x) + 1) So f'(x)=0 when : cos(x) = 0 => x = pi/2 & 3pi/2 & sin(x) = -1 => x = 3pi/2 when x < pi/2 f'(x) is negative when pi/2 < x < 3pi/2 f'(x) is positive when 3pi/2 < x < 2pi f'(x) is negative So : a) the interval on which f is increasing ( pi/2 , 3pi/2) b) the intervals on which f is decreasing [0 , pi/2) & (3pi/2 , 2pi] c) f''(x) = -14 [ cos^2(x) - sin(x) (sin(x)+1)] at x = pi/2 f''(x) is positive so there is a local minimum at x = pi/2 the local minimum = -14 at x = 3pi/2 f''(x) =0 so we study the sign of f'(x) around this value it changes from +ve to -ve so there is a local maximum at x = 3pi/2 the local maximum = 14 d) putting f''(x) = 0 [ cos^2(x) - sin(x) (sin(x)+1)] = 0 1- sin^2(x) - sin^2(x) - sin(x)=0 2 sin^(x) + sin (x) -1 = 0 (2 sin(x) -1)(sin(x) +1) = 0 sin(x) = 0.5 => x = pi/6 & 5pi/6 sin(x) = -1 => x = 3pi/2 testing the sign of f''(x) around these points we get : there are two inflection points at x = pi/6 , 5pi/6 e) the curve is concave up in the interval (0 , pi/6 ) & ( 5pi/6 , 2pi) the curve is concave down in the interval (pi/6 , 5pi/6 )