Page 83 - u4

P1: OSO/OVY
GO01962-Smith-v1.cls
UAE_Math_Grade_12_Vol_1_SE_718383_ch3
y P2: OSO/OVY QC: OSO/OVY T1: OSO July 4, 2016 13:38
See Figure 4.84 for a graph of y = f(x) over this interval. Observe that the minimum
80 value of f (the square of the distance) seems to be around 5 and seems to occur near
x = 1. We have
60
′
3
1
3
f (x) = 2(x − 3) + 4x = 4x + 2x − 6.
40
′
Notice that f (x) factors. [One way to see this is to recognize that x = 1 is a zero of
′
20 f , which makes (x − 1) a factor.] We have
′
2
x f (x) = 2(x − 1)(2x + 2x + 3).
1 2 3
FIGURE 4.84 So, x = 1 is a critical number. In fact, it’s the only critical number, since
2
2
y = (x − 3) + x 4 (2x + 2x + 3) has no zeros. (Why not?) We now need only compare the value of f at
the endpoints and the critical number. We have
f(0) = 9, f(3) = 81 and f(1) = 5.
Thus, the minimum value of f(x) is 5. This says that the minimum distance from the
√
point (3, 9) to the parabola is 5 and the closest point on the parabola is (1, 8),
which corresponds with what we expected from the graph of y = f(x).

Example 7.4 is very similar to example 7.3, except that we need to use approximate
methods to find the critical number.

y
EXAMPLE 7.4 Finding Minimum Distance Approximately
(5, 11)
2
Find the point on the parabola y = 9 − x closest to the point (5, 11). (See
9
Figure 4.85.)
(x, y)
y = 9 - x 2 Solution As in example 7.3, we want to minimize the distance from a fixed point
[in this case, the point (5, 11)] to a point (x, y) on the parabola. Using the distance
formula, the distance from any point (x, y) on the parabola to the point (5, 11) is

√
x 2 2
-4 4 d = (x − 5) + (y − 11)
√ 2 2 2
FIGURE 4.85 = (x − 5) + [(9 − x ) − 11]
y = 9 − x 2 √
= (x − 5) + (x + 2) .
2
2
2
Again, it is equivalent (and simpler) to minimize the quantity under the square root:
2
2
2
2
y f(x) = [d(x)] = (x − 5) + (x + 2) .
600 As in example 7.3, we can see from Figure 4.85 that any point on the parabola to
the left of the y-axis is farther from (5, 11) than is (0, 9). Likewise, any point on the
parabola to the right of x = 5 is farther from (5, 11) than is (5, −16). Thus, we
400 minimize f(x)for0 ≤ x ≤ 5. From the graph of y = f(x) given in Figure 4.86, the
minimum value of f seems to occur around x = 1. Next, note that
200
′
2
f (x) = 2(x − 5) + 2(x + 2)(2x)
3
x = 4x + 10x − 10.
1 2 3 4 5 Copyright © McGraw-Hill Education
′
FIGURE 4.86 Unlike in example 7.3, the expression for f (x) has no obvious factorization. Our
′
′
y = f(x) = [d(x)] 2 only choice then is to find zeros of f approximately. From the graph of y = f (x)
292 | Lesson 4-7 | Optimization