Question

f(x) = x2 (1 – In (x)) I State the intervals over which f(x) is increasing and decreasing. State any local maximum or minimum values of the function.

Answer 1

. Given that f(x) = x ² ( 3 - 1 nx) to find crctical points set of fx)=0 f' (2)= d [2? (3 - In2)] product uile za 13-152 ) + (-1,294 (22) led but search revealing = 1 (0- +43 - 194)(21) sala +(24) (3.-10%) (14)= ☆ temaniemy ..d constant)-01 aa -x+3x-2xlna f (x) = 2x - 2x10x set f (x1=0 22-2xlha=0 22nd 22 Ina = 2 2 2 int- az e a=e] at 1-2 = fle) = é? (3 - Ine) = 0 (3-)=66) i citied point (19)= (0, $j]

Domain of flie) is (0,00) we have a=e So, divide (0,6) iuto (0, e) aud (2,6) to find Interval of increase/ Decrease find of (to), xo is any point in given interval if f'(20) 30 , f (0) is increasing if 1'110) CO, f(x) is perseasing so, (0,0) of'= 210 - 2011 IN (') = 22-0= 2 >0 1 0w 2.7183 In(o)=0 (مرد) و f' (3)=213)-23)(3) = 6-61963) <o Increases ON Decrecye ON (0,0) (0,0)

working rule to find maxima and mining det f(x) be the given function stepi- find of ) Step? Enate f'(a) to zero to obtain the Stationary pointe steps? find f"G) at each stationary point (3) if 1''600) 70 then f(t) has vinimum at 4% (ii) if f1 (23)Lo theu fon) has maximum at talo lii) if t"640)=0 theu tfu may or maynot have extremuna

f" (a)= d 4'(a) = (22-24me) = 20)-(21 Ina (11) - 3 -(2+ 2nz ) = 2-2-alna f(1) = - aln(a) filles - aln(e)s -21112-230 So, at 150 Maximum occurs. Kocdl maximum at (414)=(2,ę) No Local minima