(i)   y = 7 - 5x + 6x² - 3x³

      dy/dx = -5 + 12x - 9x²

 

(ii)  y = -x + k is tangent to the curve,

        --> dy/dx = -1

      -5 + 12x - 9x² = -1

              (3x - 2)² = 0

                        x = 2/3

 

      When x = 2/3,

                  y = 7 - 5(2/3) + 6(2/3)² - 3(2/3)³ = 49/9

 

      y = k - x

      49/9 = k - 2/3

      k = 57/9

 

(iii) dy/dx = -5 + 12x - 9x²

              = -9 (x² - 12/9x + 5/9)

              = -9 (x² -4/3x + 5/9)

              = -9 [(x - 2/3)² + 1/9]

              = -9 (x - 2/3)² -1

Since -9 (x - 2/3)² < 0 and -1 < 0 for all x, dy/dx < 0

Hence, y decreases as x increases.

 

 

d/dx [x^-1(x² - 4)] = 2x.x^-1 + x^-2(x² - 4)

                        = 2 + x^-2(x² - 4)

 

Since x² - 4 > 0, x^-2 > 0, x^-2(x² - 4) > 0, 2 + x^-2(x² - 4) > 0

The function always increases as x increases.

 

 

y = (x - 3)(x + 1)^-1

dy/dx = [(x + 1) - (x - 3)]/(x + 1)²

         = 4 / (x + 1)²

 

When y = 3, x = -3, dy/dx = 1

Gradient of normal is -1.

 

Equation of normal: y - 3 = -1 (x + 3)

                                 y = -x

 

 

x + y = 9 --> x = 9 - y

 

Let L = xy²

       = (9 - y)y²

       = 9y² - y³

 

dL/dy = 18y - 3y²

 

At stationary points, dL/dy = 0

                        3y (6 - y) = 0

                        y = 0 or 6

 

d²L/dy² = 18 - 6y

 

d²L/dy²|_y=0 = 18 > 0

d²L/dy²|_y=6 = -18 < 0

 

--> Maximum value of L occurs when y = 6, x = 3.

--> Maximum value of xy² = 3 (6)² = 108