Calculus- Early Transcendentals, 2021a

5.2. Extrema of a Function 173
3. Evaluate the function at the critical values found in Step 2 and the end points of the interval.
4. Identify the absolute extrema.
Why must a function be continuous on a closed interval in order to use this theorem? Consider the
following example.
Example 5.15: Absolute Extrema of a 1/x
Find any absolute extrema for f (x)=1/x on the interval [−1,1].
Solution. The function f is not continuous at x = 0. Since 0 ∈ [−1,1], f is not continous on the closed
interval:
lim f (x)=+∞
x→0 +
lim f (x)=−∞,
x→0 −
so we are unable to apply the Extreme-Value Theorem. Therefore, f (x)=1/x does not have an absolute
maximum or an absolute minimum on [−1,1].
♣
However, if we consider the same function on an interval where it is continuous, the theorem will
apply. This is illustrated in the following example.
Example 5.16: Absolute Extrema of a 1/x
Find any absolute extrema for f (x)=1/x on the interval [1,2].
Solution. The function f is continous on the interval, so we can apply the Extreme-Value Theorem. We
begin with taking the derivative to be f ′ (x) =−1/x 2 which has a critical value at x = 0, but since this
critical value is not in [1,2] we ignore it. The only points where an extrema can occur are the endpoints
of the interval. To find the maximum or minimum we can simply evaluate the function: f (1) =1and
f (2)=1/2, so the absolute maximum is at x = 1 and the absolute minimum is at x = 2.
♣
Why must an interval be closed in order to use the above theorem? Recall the difference between open
and closed intervals. Consider a function f on the open interval (0,1). If we choose successive values of
x moving closer and closer to 1, what happens? Since 1 is not included in the interval we will not attain
exactly the value of 1. Suppose we reach a value of 0.9999 — is it possible to get closer to 1? Yes: There
are infinitely many real numbers between 0.9999 and 1. In fact, any conceivable real number close to 1
will have infinitely many real numbers between itself and 1. Now, suppose f is decreasing on (0,1): As
we approach 1, f will continue to decrease, even if the difference between successive values of f is slight.
Similarly if f is increasing on (0,1).
Consider a few more examples:
Example 5.17: Determining Absolute Extrema
Determine the absolute extrema of f (x)=x 3 − x 2 + 1 on the interval [−1,2].