A curve is defined for x>0 as y = 9 - 6x^2 - 12x^4 . a) Find dy/dx. b) Hence find the coordinates of any stationary points on the curve and classify them.

a) y = 9 -6x^2 - 12x^4 

dy/dx = -12x - 48x^3 

b) At stationary points dy/dx = 0. 

-12x - 48x^3 = 0 

48x^3 + 12x = 0 

4x^3 + x = 0

x (4x^2 + 1) = 0 

Either x = 0 or 4x^2 + 1 = 0 However, 4x^2 + 1 = 0 leads to 4x^2 = -1 which gives x^2 = -1/4. There are no solutions to this equation because x^2 is greater than or equal to 0 for all values of x. 

Therefore, the only solution is given where x = 0. When x =0, y = 9, which gives the coordinates of the stationary point as (0,9). 

In order to classify the point, we must look at the second derivative which equals: -12 - 144x^2. 

When x = 0 the second derivative = -12. As the second derivative is less than 0, the point (0,9) must be a maximum point. Therefore, the only stationary point on the curve is the maximum point (0,9).