Page 85 - How To Solve Word Problems In Calculus

EXAMPLE 1
Find the absolute maximum and minimum value of
3
f (x) = x − 12x + 5 on the interval [−1, 4]
Solution
Step1
3
f (x) = x − 12x + 5
2

f (x) = 3x − 12
2
0 = 3x − 12
12 = 3x 2

4 = x 2
x = 2 x =−2

Step2
The value x =−2 is outside the interval and is therefore
ignored.
x f (x)

−1 16
2 −11
4 21

Step3
The absolute maximum of f (x)on[−1, 4] is 21 and the
absolute minimum is −11.

Most functions that occur in connection with
maximum/minimum problems will be continuous, but
occasionally the interval representing their domain will be
open or infinite in length. In such situations, the closed
interval method fails. In fact, the function may not even have
absolute extrema.
2
The function f (x) = x on the open interval (−2, 3) has
an absolute minimum (of 0) at x = 0 but has no absolute maxi-
mum. Additional examples can easily be constructed where

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