# How do I find where the stationary points of a function are?

If you were to draw a graph of the function, a stationary point would be a point on the graph where the gradient is zero, i.e the graph has no vertical slope. For example consider the function f(x) = 2. This is a graph where every value of x simply takes the y value of 2, and thus is just the horizontal line y=2. This graph has zero gradient everywhere, and hence every point on the graph is a stationary point.

In general, if we have a function y=f(x), we must differentiate it first in order to find the stationary points. Once we have differentiated, we have an expression of the form dy/dx=f'(x). The solutions to the equation dy/dx=0 are the x values of where the stationary points occur. We then subsitute these x values into the expression y=f(x) to find the corrresponding y values to each x value. This will give us the coordinates for each stationary point.

Example

Consider the function f(x)=x^3 -12x. We let y=f(x). We must now differentiate to get an expression of the form dy/dx = f'(x). Differentiating our function with respect to x we have that f'(x)= 3x^2 - 12. Hence our expression for dy/dx is dy/dx=3x^2 - 12. We must now solve the equation dy/dx=0 in order to find the x values of the stationary points. We have 3x^2 - 12 =0 as our equations. Dividing both sides by 3, we now have x^2 - 4=0, and factorising this expression using the 'Difference of Two Squares' method, we have that (x-2)(x+2)=0. Hence our two x value are 2 and -2. When x=2, f(x)= 3(2^2)-12(2)=-12. So one coordinate is (2,-12). When x=-2, f(x)=3((-2)^2) - 12(-2) = 36. So the other coordinate is (-2,36).

Hence by differentiating y=f(x), solving the equation dy/dx=0 and then substituting in the solutions of this equation into our expression f(x), we have found that the coordinates of the stationary points are (2,-12) and (-2,36)