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Question

(i) f(x) = x4 - 62x2 + 120x + 9

(ii) f(x) = x3 - 6x2 + 9x + 15

(iii) f(x) = (x-1) (x+2)2

(iv) f(x) = 2/x-2/x2 , x>0

(v) f(x) = xex

(vi) f(x) = x/2+2/x, x>0

(vii) f(x) = (x+1) (x+2)1/3, x>-2

(viii) f(x) = x32-x2, -5<x<5

(ix) f(x) = x3-2ax2+a2x, a>0, x R

(x) f(x) = x+a2x, a>0, x ≠ 0

(xi) f(x) = x2-x2 - 2 x2

(xii) f(x) = x + 1-x, x1

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Solution

iGiven: fx = x4-62x2+120x+9f'x = 4x3-124x+120For the local maxima or minima, we must have f'x=04x3-124x+120=0x3-31x+30=0x-1x2+x-30=0x-1x+6x-5=0x=1, 5 and -6Thus, x=1, x=5 and x=-6 are the possible points of local maxima or local minima.Now,f''x=12x2-124At x=1: f''1 = 1212-124=-112<0So, x=1 is the point of local maximum.The local maximum value is given byf1 = 14-6212+120×1+9=68At x=5: f''5 = 1252-124=176>0So, x=5 is the point of local minimum.The local minimum value is given byf5 = 54-6252+120×5+9=-316At x=-6: f''-6 = 12-62-124=308>0So, x=-6 is the point of local maximum.The local minimum value is given by f-6 =-64-62-62+120×-6+9=-1647


iiGiven: fx = x3-6x2+9x+15f'x = 3x2-12x+9For the local maxima or minima, we must have f'x=0 3x2-12x+9=0 x2-4x+3=0x-1x-3=0x=1 and 3Thus, x=1 and x=3 are the possible points of local maxima or local minima.Now, f''x = 6x-12At x=1: f''1 = 61-12=-6<0So, x=1 is the point of local maximum.The local maximum value is given byf1 = 13-612+9×1+15=19At x=3: f''3 = 63-12=6>0So, x=3 is the point of local minimum.The local minimum value is given byf3 =33-632+9×3+15=15


iii Given:fx =x-1x+22=x-1x2+4x+4=x3+4x2+4x-x2-4x-4=x3+3x2-4f'x = 3x2+6xFor the local maxima or minima, we must have f'x=0 3x2+6x=03xx+2=0x=0 and-2Thus, x=0 and x=-2 are the possible points of local maxima or local minima.Now, f''x = 6x+6At x=0: f''0 = 60+6=6>0So, x=0 is the point of local minimum.The local minimum value is given byf0 = 0-10+22=-4At x=-2: f''-2 = 6-2+6=-6<0So, x=-2 is the point of local maximum.The local maximum value is given by f-2 =-2-1-2+22=0


ivGiven: fx = 2x-2x2=2x-1-2x-2f'x =-2x-2+4x-3=4x3-2x2For the local maxima or minima, we must have f'x=04x3-2x2=04-2x=0x=2Thus, x=2 is the possible point of local maxima or local minima.Now, f''x = -12x4+4x3At x=2: f''2 =-1216+48 =-12+816=-14<0So, x=2 is the point of local maximum.The local maximum value is given byf2 = 22-222=1-12=12


v Given:fx = xexf'x = ex+xexFor the local maxima or minima, we must have f'x=0ex+xex=0ex1+x=0ex0 , x=-1x=-1Thus, x=-1 is the possible point of local maxima or local minima.Now,f''x = ex+ex+xexAt x=-1: f''-1 =e-1+e-1-e-1 =e-1>0So, x=-1 is the point of local minimum.The local minimum value is given byf-1 = -e-1=-1e

vi Given:fx = x2+2xf'x = 12-2x2For the local maxima or minima, we must have f'x=012-2x2=0x2=4x=2 and-2Thus, x=2 and x=-2 are the possible points of local maxima or a local minima.Since x>0, x=2Now, f''x = 4x3At x=2: f''2 =423 =12>0So, x=2 is the point of local minimum.The local minimum value is given by f2 = x2+2x =1+1=2

vii Given:fx =x+1x+213f'x =x+213+13x+1x+2-23For the local maxima or minima, we must have f'x=0 x+213+13x+1x+2-23=013x+1=-x+213×x+22313x+1=-x+2x+1=-3x-6x=-74Thus, x=-74 is the possible point of local maxima or local minima.Now, f''-74 =23x+2-23-29x+1x+2-53At x=-74: f''-74 =23-74+2-23-29-74+1-74+2-53=2314-23+11814-52>0So, x=-74 is the point of local minimum.The local minimum value is given byf-74 = -74+1-74+213=-341413=-3443


viii Given: fx = x32-x2f'x = 32-x2-x232-x2For the local maxima or minima, we must have f'x=032-x2-x232-x2=032-x2=x32-x232-x2 =x2 x2=16 x=±4 Thus, x=4 and x=-4 are the possible points of local maxima or local minima.Now, f''x =-x32-x2 -2x32-x2+x332-x232-x2=-x32-x2 -2x32-x2+x332-x232-x2At x=4: f''4 = -432-42 -832-42+4332-4232-42 =-1-19264=-3<0So, x=4 is the point of local maximum.The local maximum value is given by f4 = 432-42 =16At x=-4: f''-4 = 432-42 +832-42-4332-4232-42 =1+2=3>0So, x=-4 is the point of local minimum.The local minimum value is given by f-4 = -432-42 =-16


ix fx = x3-2ax2+a2xf'x = 3x2-4ax+a2For the local maxima or minima, we must have f'x=03x2-4ax+a2=0 3x2-3ax-ax+a2=03xx-a-ax-a=03x-ax-a=0x=a and a3Thus, x=a and x=a3 are the possible points of local maxima or local minima.Now, f''x = 6x-4aAt x=a: f''a = 6a-4a=2a>0So, x=a is the point of local minimum.The local minimum value is given byfa = a3-2aa2+a2a=0At x=a3: f''a3 = 6a3-4a=-2a<0So, x=a3 is the point of local maximum.The local maximum value is given byfa3 =a33 -2aa32 +a2a3 =a327-2a39+a33=4a327


xGiven: fx = x+a2xf'x = 1-a2x2For the local maxima or minima, we must havef'x=0 1-a2x2=0x2=a2x=±a Thus, x=a and x=-a are the possible points of local maxima or local minima.Now, f''x = a2x3At x=a: f''a =a2a3 =1a>0So, x=a is the point of local minimum.The local minimum value is given byfa=x+a2x=a+a=2aAt x=-a: f''a =a2-a3 =-1a<0So, x=-a is the point of local maximum.The local maximum value is given byf-a = x+a2x =-a-a=-2a


xiGiven: fx = x2-x2f'x = 2-x2-x22-x2For the local maxima or minima, we must have f'x=02-x2-x22-x2=02-x2=x2-x22-x2 =x2 x2=1 x=±1 Thus, x=1 and x=-1 are the possible points of local maxima or local minima.Now, f''x =-x2-x2 -2x2-x2+x32-x22-x2=-x2-x2 -2x2-x2+x32-x22-x2At x=1: f''1 = -12-12 -22-12+132-122-12 =-12-32=-2<0So, x=1 is the point of local maximum.The local maximum value is given byf4 = 12-12 =1At x=-1: f''-1 = 12-12 +22-12-132-122-12 =1+1=2>0So, x=-1 is the point of local minimum.The local minimum value is given byf-1 = -12-12 =-1


xii Given: fx = x+1-xf'x = 1-121-xFor the local maxima or minima, we must have f'x=01-121-x=01-x=121-x=14 x=34 Thus, x=34 is the possible point of local maxima or local minima.Now, f''x = -141-x41-xAt x=34: f''34 = -141-3441-34 =-12<0So, x=34 is the point of local maximum.The local maximum value is given byf34 = 34+1-34 =54

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