What's the correct answer for c and d

17. Consider the function f(x) = x -x-11 . a. Find the average rate of change of f(x) on [0,6]. (2 points) 2 f (0 ) = (0 ) 2 - 0 - 11 =-110* f ( 6 ) = 162 -6-11=19 f ( 6 0 ) - f (o ) the average rate of 6 - 0 change 18 5 = 19 - (-11) = 30 - 5 b. Find the instantaneous rate of change of f (x) when x = 4. 6 - 0 fl (x ) = 2x - 1 (2 points) 2 f ) ( x ) = 2 ( 4 ) - 1 ( ( 4 ) = 8- 1 c. Write an equation of the line tangent to f at the point where x = 4. (2 points) f ( 4 ) = ( 4)8 - 4-11 =1 (4, 17 f ? ( x ) = 2 x - 1 f' ( ) = ! ) a value ( number ) s should be the slope y = m ( x - x ) + y! y= (2x-1 (x-4) + 1/ because at the point ( 4, # 1 ) there is a Specific slope d. Still using f(x) above, if g(x) = J (x) ) It s not variable find all values of x where the tangent lines to g are horizontal. Show how you arrive at your answer. (3 points) 0= 2x- 1 Ox - 1 11 = xax x 9 ( x ) = (ex ]( f? ( x ) ) - ( f ( x ) ) . (ex ) 1 = ax ex = ex( 2x - 1 ) - (ex ( x8- x - 11 ) when X equals ex 6 : 3X- - xat x +ll. a number ( ex ) greaterthan 3 the valve 2x - 1 exya X +llex 2x - 1 - ex (xo - x - 1 1) of the ex sope gets ex ( 3)0- 3 - 11 close to 9- 3 - 11