State the domain and range of each function below. Explain your process. [A4] [C2] Function Reasoning f(x) = _ 13 √− 4x?+ 16+ 19 f(x) = _1 6x?+ ?81− 99 5. Calculate the average rate of change for the function f(t) = 2 x 3 + 5 over the interval −1 ≤ x ≤ 2. [K3] 6. a. Use an x− interval of 0.0001 to determine the instantaneous range of change of the function f(x)= x 4 +  8 x 3  + 19 x 2 + 12x at x = −2. Give answer to the nearest tenth. [A3] b. Determine whether the function f(x)= x 4 +  8 x 3  + 19 x 2 + 12x at x = −2 has a maximum or minimum at the point where x = −2. Justify your answer. 

 f() = 13)-42+ 16+ 19. since square root is possible for only non-negative quantities , we must have -4x+16>0.3x<4 thes Domain ( f) = { nE RR: 2 547= (-0, 4] As 7 - - 08, f ( x) - tos. Further , f (4 ) = 19 . Again since for each - E Domain (f ), we must have f (a) > 19 . Since Domain (f) is connected and f is continuous on domain (f ) , we must have [ Domain (f ) ] = Range (f ) is also connected . Thes, Range ( ) = [19, 00). f ( x ) = Txt 81 - 99. since division by Zero is not possible , we must have 62 + 81 7 0 > n7 - 21 7 27 - 27 so, Domain ( f) = MR . {- 27] we claim that Range ( f ) = 1R . {-39] . Let CE TRAS- 997. Then ( * - 99. => C + 99 #0 Let no egg - 81) /6 . Then f ( x. ) = G 6 xo + 81 - 99 = C . - C F Range (f ). Again if c=gg then ce Range (f ) For otherwise , there exists yf IR such that , fly )- c=-ss 6ytl = 0, which is not possible . The Range (f ) = IR - 1- gg].