What are the maxima and minima?
The maxima of a function f(x) are all the points on the graph of the function which are 'local maximums'. A point where x=a is a local maximum if, when we move a small amount to the left (points with xa), the value of f(x) decreases. We can visualise this as our graph having the peak of a 'hill' at x=a.
Similarly, the minima of f(x) are the points for which, when we move a small amount to the left or right, the value of f(x) increases. We call these points 'local minimums', and we can visualise them as the bottom of a 'trough' in our graph.
One similarity between the maxima and minima of our function is that the gradient of our graph is always equal to 0 at all of these points; at the very top of the peaks and the very bottom of the troughs, the slope of our graph is completely flat. This means our derivative, f '(x), is equal to zero at these points.
How do we find them?
1) Given f(x), we differentiate once to find f '(x).
2) Set f '(x)=0 and solve for x. Using our above observation, the x values we find are the 'x-coordinates' of our maxima and minima.
3) Substitute these x-values back into f(x). This gives the corresponding 'y-coordinates' of our maxima and minima.
Which of these points are maxima and which are minima?
Here we may apply a simple test. Assume we've found a stationary point (a,b):
1) Differentiate f '(x) once more to give f ''(x), the second derivative.
2) Calculate f ''(a). If f ''(a)<0 then (a,b) is a local maximum.
If f ''(a)>0 then (a,b) is a local minimum.
To see why this works, imagine moving gradually towards our point (a,b), plotting the slope of our graph as we move. If our point is a local maximum, we can that this slope starts off positive, decreases to zero at the point, then becomes negative as we move through and past the point. Our slope, f '(x), is decreasing throughout this movement, so we must have that f ''(a)<0.
The exact reverse is true if (a,b) is a local minimum. Our slope is increasing through the same movement, so here we have that f ''(a)>0.
Find the maxima and minima of f(x)=x3+x2.
First, we find f '(x). Using the rules of differentiation, we find f '(x)=3x2+2x.
Now let's set f '(x)=0: 3x2+2x=0
x=0 or x=-2/3
Substitute these values back in so that we can find our 'y-coordinates': f(0)=(0)3+(0)2=0
Hence our stationary points are (0,0) and (-2/3,4/27).
Finally, we use our test: f ''(x)=6x+2
f ''(0)=2 (substituting x=0)
f ''(-2/3)=-2 (substituting x=-2/3)
2>0, so (0,0) is a local minimum of f(x).
-2<0, so (-2/3,4/27) is a local maximum of f(x).