Find the stationary points of the curve y=x^4-8x^2+3 and determine their nature.

First we will differentiate y to find dy/dx= 4x³-16x. Stationary points exist at dy/dx=0 so we set 4x³-16x=0 and solve for x. x(4x²-16)=0. So x=0 or 4x²-16=0. If 4x²-16=0 then x²=4 and so x=2 or x=-2. By subbing in our values of x=0, 2, -2 into our original equation we can find the corresponding value of y for each of these x values. They give y=3, -13, -13 respectively. Hence we have found that there are three stationary points at (0, 3), (2, -13), (-2, -13). To determine the nature of these stationary points we need to find the second derivative of the curve, which is to differentiate dy/dx. This gives, d²y/dx²= 12x²-16. Now we sub in the values of x at each of our stationary points to determine their nature. At x=0, d²y/dx²= -16 which is less than 0 so the stationary point is a maximum. At x=2, d²y/dx²= 32 which is greater than 0 so is a minimum. At x=-2, d²y/dx²= 32 which again is greater than 0 and so is a minimum. And so here we have found all stationary points and determined their nature as required.