MAXIMA AND MINIMA
87
X
/to
0
-4
*0
5
*i
-5
2
-4
13.34 Find the absolute maximum and minimum of
on
Clearly, the absolute minimum is 0, attained where cosx=j, that is, at x = irl3 and x = 5ir/3. So,
we only have to determine the absolute maximum. Now,
(—sin jc). We need only consider
the critical numbers that are solutions of sin x = 0 (since the solutions of cos x = \ give the absolute
minimum). Thus, the critical numbers are 0, TT, and 2ir. Tabulation of/(*) for these numbers shows that the
absolute maximum is |, achieved at x = IT.
X
/to
0
i
2
7T
3
27T
1
2
13.35
Find the absolute maximum and minimum (if they exist) of f(x) = (x + 2)l(x - 1).
Since lira f(x) = +» and lim f(x) = -<*>, no absolute maximum or minimum exists.
v^l +
X—»1
13.36
Find the absolute maximum and minimum of
on
Since f(x) is not differentiable at x = f (because |4x - 3j is not differentiable at this point), x = |
is a critical number.
(Here, we have used the chain rule and Problem
9.47.) Thus, there are no other critical numbers. We need only compute f(x) at x = i and at the endpoints.
We see that the absolute maximum is V5, attained at x = 0, and the absolute minimum is 0, attained at
v = 2
X — 4 .
X
/to
0
V5
1
0
1
1