First of all you need to establish whether the given function on the whole interval [a, b] and if it has break points, what kind of these breaks. For example, the function f(x) = 1/x does not have neither maximum nor minimum values on the interval [-1, 1] because the point x = 0 tends to plus infinity on the right and negative infinity on the left.
If the given function is linear, i.e. given by the equation of the form y = kx + b, where k ≠ 0, it is in all its scope is monotone increasing if k > 0; and decreases monotonically if k < 0. Therefore, its maximum value for any given segment will be f(b), if k > 0; and f(a), if k < 0.
The next step is the study of the function at the extrema. Even if it is determined that f(a) > f(b) (or Vice versa), the function may reach a large value at the point of max.
To find the maximum point, it is necessary to resort to the help of the derivative. We know that if at point x0 the function f(x) has a extremum (i.e. maximum, minimum or stationary point), then its derivative f'(x) at that point vanishes: f'(x0) = 0.

To determine which of three kinds of extremum is detected point, it is necessary to investigate the behavior of the derivative in its vicinity. If it changes sign from plus to minus, that is, monotonically decreasing, then the point of the original function has a maximum. If the derivative changes sign from minus to plus, i.e., monotonically increases, at the point of the original function has a minimum. If, finally, the derivative does not change sign, then x0 is a stationary point for the original function.
In cases where to compute the signs of the derivative in the vicinity of the point difficult, you can use the second derivative f"(x) and determine the sign of this function at the point x0:

- if f"(x0) > 0, the found minimum point;
- if f"(x0) < 0, the found maximum point;
finally, if f"(x0) = 0, then the stationary point.
For the final solution of the problem is necessary to select the maximum of the values of the function f(x) at the endpoints of the interval and all found high.