Find the minimum and maximum points of the graph y = x^3 - 4x^2 + 4x +3 in the range 0<=x <= 5.

First, use the standard method of setting the derivative to be equal to 0 to find the stationary points. This yields the equation 3x^2 - 8x + 4 = 0 and so the stationary points are at x = 2/3 and 2 respectively.It is possible to then continue by finding the second derivative to determine the nature of each turning point, but this is not necessary when considering the general shape of an x^3 graph. The maximum comes before the minimum and so there is a local maximum at x=2/3 and a local minimum at x=2. A careful check that the two stationary points are not both points of inflection can be done by just comparing the values of the graph at the two stationary points and seeing that the value decreases from x=2/3 to x=2, and so we indeed have a maximum and then a minimum.
The only values where we can have a maximum or minimum over the interval in the question are at the stationary points, or at the boundaries (this is often forgotten at A-level) x=0 and x=5. So, calculating the values at each of these four points gives:x=0, y=3 x=2/3, y = 113/27 x=2,y=3 x=5,y=48So the final answer is that there is a minimum at (0,3) or (2,3) ( both answers would be acceptable) and a maximum at (5,48).