The equation of a line is y=3x – x^3 a) Find the coordinates of the stationary points in this curve, stating whether they are maximum or minimum points b) Find the gradient of a tangent to that curve at the point (2,4)

a) A stationary point is any point on the curve that is flat, still, not increasing or decreasing. Another way to think of this is the gradient at a stationary point = 0

Firstly make an equation for your gradient, differentiating y=3x – x³ gives dy/dx = 3 – 3x²

In order to find the coordinates, we must find values of x and y when the gradient = 0. Solving the derivative equation above to find x,

3 – 3x² = 0

3=3x²

x²=1

x= +1 or -1

Now find the corresponding y values by substituting x into the original equation:

When x = 1, y = 3(1) – (1)³ y= 2

when x= -1, y = 3(-1) – (-1)³ y= -2

To find whether each of these points is a minimum or a maximum point, you must differentiate the original equation twice: y=3x – x³

dy/dx = 3 – 3x²

d²y/dx² = -6x

when x=1, d²y/dx² = -6(1) d²y/dx² = -6 -6 is <0, therefore it is a maximum point

when x=-1, d²y/dx² = -6(-1) d²y/dx² = 6 6 is >0, therefore it is a minimum point

So (1,2) is a maximum stationary point and (-1,-2) is a minimum stationary point

b) Note: remember that the gradient of a curve at a point is also the gradient of the tangent to the curve at that point.

We already know the equation for the gradient function of the curve, dy/dx = 3 – 3x² So to find the gradient at the point (2,4) simply substitute x = 2 into the equation: dy/dx = 3 – 3x²

= 3 – 3(2)²

= -9