Question

Find any inflection points and the intervals of concavity for the function y = 3x-1 x2

Answer 1

and concavity : where occurs Inflection points Inflection second derivative concarita Changes. points j.e.) where the zero 2 Concave upward A function is Said to be interval it " >0 at each point on the interval if f'(X) co at each point Concave downward interval on 33 ) - 3x-1 Y = X² X2 x2 1-42 lu) = It is lxh) = nxn- 3 d y' = MLX --*2) Ir x CS Scanned with CamScanner

=-36-2) 3-2-1 E =36) 3 (-2) x-2-1 +2x-3 >> 4- =-30-2 +2x-3 y" d (Y) + 2x-3 2 d da Р P 34-2+2x+3] (- +2 6 6 .4 -3 6 x -6.3 X4 6 yli 60-6 214 62-6 'yl=0 =) =0=) 676-6=0->) 6x=6 24 n=1 - occurs at Inflection K 64-6 *4 point Test for Some values of x=-1,0.5, 2 in yl, a y"(-) = 6(-1)-6 -6-65 (-1) 4 y"(-1) <0 -12 E (0.5)4 y" (0.5)= 6(05) -6 (0.5)4 Layllo:) so Y" (2) = 6(2)-6 = (294 .:"(2) >0 6 91 y(x) 30 y"(x) <0 yl (x) <0 CS Scanned with CamScanner

x=1 is a point, because inflection concarity Changes evaluations from above Hence 30-1 2 32-1 = y(i)= Y- x2 2 cr (1) Inflection point is (1,2) Concare upward on (176) concare downward on 600,oluCO,) CS Scanned with CamScanner