Given that ' x = —(x -305") f ( ) (:c—2)2(auc2+1)3\/;u:_2 is horizontal

We find the values of x where tangent line is horizontal.

Step-by-step explanation

we know that, when the derivative is zero then tangent line is horizontal. Herre , derivative of function is given by F ' ( x ) = ( 2 " - 2 ) ( x+1 ) ( x - 2 ) 2 ( x 2 +1 ) 8 8 2 Assume f ( x ) = 0 0 = (2 2- 2 ) ( x + 1 ) (2 -2) 2 ( 212 + 1 ) 31 x 2 : ( 2 " - 2 ) ( x + 1 ) = 0 [ ( 2 -2)" ( x2 +1 ) 3/ 2 2 # 0 ] ble know that product of two terms is zew if and only if one of them is zero . 2 2 - 2 = 0 a + 1 = 0 a = ! 21 = - J a = ( + ( 2 ) - 0 x = -1 [: a = + Vaz = (+ Ja)= ] a = tvz or a = - 1 . The values of a where tangent line is horizontal are 2 = 12, a = - V2 x = - 1 . : Required values of a are V2, - 12, - 1.