Because it was not explained to me. I understand all the other points you have made except this one. Go on and start from scratch.
Starting from scratch.
Your
ORIGINAL problem concerned finding the range of a real function defined over a closed domain. What is the range of f(x) defined over [a, b]?
Do you agree that was the general problem? Do you understand the problem? You can scarcely understand the answer if you do not understand the problem.
It was pointed out to you that such problems can easily be solved, at least approximately, with a graphing calculator. No muss, no fuss.
Do you understand that? You, however, wanted to know how to solve such problems analytically.
If f(x) is differentiable over (a, b), an obvious analytic method is to locate the extrema using calculus, calculate the value of the function at each extremum and at a and b, and list the minimum and maximum of the results.
Is there anything you do not understand about that? If so, where exactly are you lost?
Suppose f(x) is not differentiable over (a, b) but can be redefined into differentiable pieces. For example, a < c < b, g(x) is differentiable over (a, c), g(x) [imath]\equiv[/imath] f(x) over [a, c], h(x) is differentiable over (c, b), and h(x) [imath]\equiv[/imath] f(x) over [c, b]. For that example, calculate the range for g(x) over [a, c] and the range of h(x) over [c, b] and then take the minimum of the two ranges as the minimum for the range of f(x) and the maximum of the two ranges as the maximum for the range of f(x). You can generalize this method to as many pieces as are needed to find differentiable equivalents to f(x) over the entire interval of [a, b].
Do you understand fully? if not, where precisely do you lose track.
Now in fact, the set of exercises that you were asked to work on were far less general. They consisted of functions of the following form:
[math]f(x) = u(x) + |v(x)|, \text { over } [a, \ b].\\
\therefore \ f(x) = u(x) - v(x) \text { if } x \text { such that } v(x) \le 0.\\
\text { And } f(x) = u(x) + v(x) \text { if } x \text { such that } v(x) \ge 0.[/math]
Moreover, u(x) and v(x) were differentiable but linear and thus had no internal extrema. Therefore, f(x) was decomposable into at most two linear pieces with extremes at the boundaries. To find the boundaries of the pieces, all that was required was to find the zero of v(x) where v(x) is a linear function.
Where is your difficulty in understanding?
