Chapter 4 | Applications of Derivatives
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Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous functions. Example 4.13 Locating Absolute Extrema
For each of the following functions, find the absolute maximum and absolute minimum over the specified interval and state where those values occur.
a. f ( x ) =− x 2 +3 x −2 over [1, 3]. b. f ( x ) = x 2 −3 x 2/3 over [0, 2].
Solution a. Step 1. Evaluate f at the endpoints x =1 and x =3.
f (1) =0and f (3) =−2
Step 2. Since f ′( x ) =−2 x +3, f ′ is defined for all real numbers x . Therefore, there are no critical points where the derivative is undefined. It remains to check where f ′( x ) =0. Since f ′( x ) =−2 x +3=0 at x = 3 2 and 3 2 is in the interval [1, 3], f ⎛ ⎝ 3 2 ⎞ ⎠ is a candidate for an absolute extremum of f over [1, 3]. We evaluate f ⎛ ⎝ 3 2 ⎞ ⎠ and find f ⎛ ⎝ 3 2 ⎞ ⎠ = 1 4 .
Step 3. We set up the following table to compare the values found in steps 1 and 2. x f ( x ) Conclusion
0
0
3 2
1 4
Absolute maximum
3
−2
Absolute minimum
From the table, we find that the absolute maximum of f over the interval [1, 3] is 1 4 ,
and it occurs at
x = 3 2 . The absolute minimum of f over the interval [1, 3] is −2, and it occurs at x =3 as shown in the following graph.
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