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).
