Question

a) Verify the Rolle's theorem for the function f(x) = -1 x +x-6 over the interval (-3, 2] 3-X b) Find the absolute maximum and minimum values of function f(x)= (1+x?)Ě over the interval [-1,1] c) Find the following for the function f(x) = 2x – 3x – 12x +8 i) Intervals where f(x) is increasing and decreasing. ii) Local minimum and local maximum of f(x) iii) Intervals where f(x) is concave up and concave down. iv) Inflection point(s). v) Sketch the curve f(x) Question No: 4 Marks: 1 + 2+ (2 + 2)+2 dy: y = (sin x)" dx a) Use logarithmic differentiation to find b) Find }(Vx+3x+ 10]dx c) If f'(x) = sin x–cosx and the curve f(x) passes through the point (3:2) i) Find f(x) ii) Find the local extreme (if any) of f(x) using second derivative test. d) Evaluate using integration by parts: [x cos 3x dx

Answer 1

® a) for f(x) = x²+x-6 Lover the interval [-3,2] 3-2 There are three conditions for Roll's theorem f(x) Continuous at (-3,2) Hence this function is continuons Because at every over than which not [-32] at (-3,2) 2 3 derivable on (-3,2) because f'(x) = (3-2) (2x+1) (x²+x-6) (-1) (3-2) 2 exists au x 7 3 which does lie en (-3,2) not 3) f(-3) f (2) 4) Hence there exist CE[-3,2] showing f'(c)=0 ( 3-x) (2x+1) + (x+3=6+ de (3-2) 2 then f' (c) O (3-6) (20+) + (246-6) = 0 6c+3-2c²-4tc²+4-6=0 c² + 6c c² – 6c + 3=0 CE3+56 4 c=3-56 then = 0.55 E[-3,2] - 3-o C

Date b for f(x) = (1+x/ (1-x2) 73 (-22) over maxima & minina absolute f'(x) = 2 3 put 2. (1-x23 -2x)=0 (1-2y3 to then (-20) = 0 220 Cao Mow Check for x= f(-) 1 Absolute minima Absolute maxima € Absolute minima 2x3 3x². - 122 + 8 f'(x) = 6x² - 6x-12 put f'(x)=0 for critical points 6x² - 6x-12 = 0 2 x x ² - x - 2 = 0 -20 te-2=0 x(x-2) +2x-2) x= -1, 2 finding f'(x) = 12x - 6 f"(-1) = -18 <0 then for f" (2). = 18 30 x = -1 a local maxima is a local วๆ 174 then x = 2

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