Calculus- Early Transcendentals, 2021a

174 Applications of Derivatives
Solution. First, notice f is continuous on the closed interval [−1,2], so we’re able to use Theorem 5.14 to
determine the absolute extrema. The derivative is f ′ (x)=3x 2 − 2x, and the critical values are x = 0,2/3
which are both in the interval [−1,2]. In order to find the absolute extrema, we must consider all critical
values that lie within the interval (that is, in (−1,2)) and the endpoints of the interval.
f (−1)=(−1) 3 − (−1) 2 + 1 = −1
f (0)=(0) 3 − (0) 2 + 1 = 1
f (2/3)=(2/3) 3 − (2/3) 2 + 1 = 23/27
f (2)=(2) 3 − (2) 2 + 1 = 5
The absolute maximum is at (2,5) and the absolute minimum is at (−1,−1).
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Example 5.18: Determining Absolute Extrema
Determine the absolute extrema of f (x)=−9/x − x + 10 on the interval [2,6].
Solution. First, notice f is continuous on the closed interval [2,6], so we’re able to use Theorem 5.14 to
determine the absolute extrema. The function is not continuous at x = 0, but we can ignore this fact since
0 is not in [2,6]. The derivative is f ′ (x)=9/x 2 − 1, and the critical values are x = ±3, but only x =+3is
in the interval. In order to find the absolute extrema, we must consider all critical values that lie within the
interval and the endpoints of the interval.
f (2)=−9/(2) − (2)+10 = 7/2 = 3.5
f (3)=−9/(3) − (3)+10 = 4
f (6)=−9/(6) − (6)+10 = 5/2 = 2.5
The absolute maximum is at (3,4) and the absolute minimum is at (6,2.5).
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When we are trying to find the absolute extrema of a function on an open interval, we cannot use
the Extreme Value Theorem. However, if the function is continuous on the interval, many of the same
ideas apply. In particular, if an absolute extremum exists, it must also be a local extremum. In addition
to checking values at the local extrema, we must check the behaviour of the function as it approaches the
ends of the interval.
Some examples to illustrate this method.
Example 5.19: Extrema of Secant
Find the extrema of sec(x) on (−π/2,π/2).
Solution. Notice sec(x) is continuous on (−π/2,π/2) and has one local minimum at 0. Also
lim sec(x)= lim sec(x)=+∞,
x→(−π/2) + x→(π/2) −
so sec(x) has no absolute maximum, but the point (0,1) is the absolute minimum.
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