Question

1. If a particle moves according to a law of motion S(t)=12-6-7, t 20 Where t is measured in seconds and sin meters, (a) Find the velocity of the particle in terms of t. (b) Find the velocity and the speed at time t=1. (c) When is the particle at rest? (d) When is the particle moving to the right and when is it moving to the left? (e) Find the acceleration of the particle at t. (10pts) 2. Evaluate the following limits if they exist: (a) lim ++5x+4 x-14 x2 +38-4 (b) lim 72+27-8 (c) lim -2 (15pts) 3. Find an equation of the tangent line to the curve y = x2-3 at the point (1,-2) (10pts) 4. Differentiate i.e. find X (a) y +Vx3 +2e* (b) y = 3x22 (c)y = (x3 +x-1) (x6 -7x2 +3 5. Use the Squeeze theorem to show that lim f(x) = 7 if 4x-95 f(x) < x2-4x+7 (15pts) (5pts) 6. Explain what is meant to say that lim lim f(x) exists? Explain. f(x) = 3 and lim f(x)=7. In this situation is it possible that (10pts) 20 h Xa xa 7. Use either definition f'(a) = limf(a+h)-f(a) or f'(a) = lim" (=f@ to find the derivative of f(x)= 5x - 1 at x=2 (i.e find f '(2)) (10pts) 8. Find the domain of (a) f(x) = 2 - 6x +5 (b) g(x) =*=52 (10pts) 9. Given f(x) = 2xe* find (a) f'(x) (b) f"(x) (c)f"(0) (15pts) x2+2x-7

Answer 1

29 a) /'m +5 + 4 or 1) Position of particle is axa4 x² + 3x - 4 sit)= 4² t - 6t it to a) velocity & I lim (x+4)(x+1)-(4h (4+1) 40 rets=5 ct) a n da koh (x+4)(x-1) vet)=2t -6 (gth - ② b) At time t=1 V(1) = 2011 –6 = -4 m/s. b tim x - x = lim 1-12) speed = luci) = 1-41=14 mis. to x²+27-8 a 200 x1 + 2 + 8) c) Partical At rest of ult=0 → 26-6=0 lt=3) sec c tim I - im d) partide moving to right vlt) To - C) no - 1-x² w i- (1-x)(1+2) particle moving toleft VCA) LO 2t - 670 = t > 6 7+73 - (1-1)2 = 0 moving right : t73 3 y=x²-3 = dy = 22 Moving left :12 <3 at (1,-2) a dylo slope of T = 2 e) Acch = a (uct)) = a (21-6) Gan of tungent i y-y, amrare) facch. = 2 802/5 | - 4-12) = 2(0-1) =) y +2 =2x-2 y=20-4 ar dal, 2)

29 a) /'m +5 + 4 or 1) Position of particle is axa4 x² + 3x - 4 sit)= 4² t - 6t it to a) velocity & I lim (x+4)(x+1)-(4h (4+1) 40 rets=5 ct) a n da koh (x+4)(x-1) vet)=2t -6 (gth - ② b) At time t=1 V(1) = 2011 –6 = -4 m/s. b tim x - x = lim 1-12) speed = luci) = 1-41=14 mis. to x²+27-8 a 200 x1 + 2 + 8) c) Partical At rest of ult=0 → 26-6=0 lt=3) sec c tim I - im d) partide moving to right vlt) To - C) no - 1-x² w i- (1-x)(1+2) particle moving toleft VCA) LO 2t - 670 = t > 6 7+73 - (1-1)2 = 0 moving right : t73 3 y=x²-3 = dy = 22 Moving left :12 <3 at (1,-2) a dylo slope of T = 2 e) Acch = a (uct)) = a (21-6) Gan of tungent i y-y, amrare) facch. = 2 802/5 | - 4-12) = 2(0-1) =) y +2 =2x-2 y=20-4 ar dal, 2)