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                                   GO01962-Smith-v1.cls
    UAE-Math-Grade-12-Vol-1-SE-718383-ch4
                    CHAPTER 4 • •
                                  Applications of Differentiation
             262  P2: OSO/OVY  QC: OSO/OVY  T1: OSO   October 25, 2018  17:24                                  4-52
                         y
                                          EXAMPLE 6.3     A Graph with Two Vertical Asymptotes
                      200                                       2
                                          Draw a graph of f(x) =  x  showing all significant features.
                                                               2
                      150                                     x − 4
                      100                 Solution The default graph produced by our computer algebra system is seen in
                                          Figure 4.68a, while the default graph drawn by most graphing calculators looks like
                       50                 the graph seen in Figure 4.68b. Notice that the domain of f includes all x except
                                          x =±2 (since the denominator is zero at x =±2). Figure 4.68b suggests that there
                                      x
              -4                   4      are vertical asymptotes at x =±2, but let’s establish this carefully. We have
                      -50
                                                                                  +
                                                                   x 2            x 2
                    FIGURE 4.68a                             x→2 x − 4  = lim +          =∞.                (6.7)
                                                             lim
                                                                  2
                                                                         x→2 (x − 2)(x + 2)
                                                               +
                      y =  x 2                                                 +     +
                           2
                          x − 4
                         y                Similarly, we get
                                                            lim   x 2  = −∞,   lim   x 2  = −∞              (6.8)
                                                                 2
                                                                                    2
                        5                                   x→2 x − 4         x→−2 x − 4
                                                                                  +
                                                               −
                                       x
             -10   -5         5   10                                        x 2
                      -5                  and                         lim       =∞.                         (6.9)
                                                                           2
                                                                         −
                                                                     x→−2 x − 4
                                          Thus, there are vertical asymptotes at x =±2. Next, we have
                    FIGURE 4.68b
                      y =  x 2                                        2       2
                           2
                          x − 4                              f (x) =  2x(x − 4) − x (2x)  =  −8x  .
                                                              ′
                                                                         2
                                                                                       2
                                                                       (x − 4) 2     (x − 4) 2
                                          Since the denominator is positive for x ≠ ±2, it is a simple matter to see that
                                                             { > 0, on (−∞, −2) and (−2, 0)  f increasing.
                                                          ′
                                                         f (x)                                             (6.10)
                                                               < 0, on (0, 2) and (2, ∞).  f decreasing.
                                          In particular, notice that the only critical number is x = 0 (since x =−2, 2 are not in
                                          the domain of f). Thus, the only local extremum is the local maximum located at
                                          x = 0. Next, we have

                                                                          2
                                                                 2
                                                             2
                                                                               1
                                                   ′′
                                                  f (x) =  −8(x − 4) + (8x)2(x − 4) (2x)  Quotient rule.
                                                                    2
                                                                  (x − 4) 4
                                                                   2
                                                           2
                                                                            2
                                                         8(x − 4)[−(x − 4) + 4x ]
                                                                                             2
                   -       0   +                      =                              Factor out 8(x − 4).
                                                                 2
                                   (x - 2) 3                    (x − 4) 4
                           2
                                                            2
               -   0       +                             8(3x + 4)
                                   (x + 2) 3          =                              Combine terms.
                                                           2
                  -2                                     (x − 4) 3
               +   ×   -   ×   +
                                                              2
                                   f��(x)                  8(3x + 4)
                  -2       2                          =              .               Factor difference of two squares.
                                                              3
                                                         (x − 2) (x + 2) 3
         Copyright © McGraw-Hill Education   +  -2 × ×  +  - 0 0  -  × 2 2  -  f�(x)  denominator, as seen in the margin. We then have  Concave up.  (6.11)
                                          Since the numerator is positive for all x, we need only consider the factors in the
                                                                > 0, on (−∞, −2) ∪ (2, ∞)
                                                              {
                              ×
               +
                                +
                                                          ′′
                                                         f (x)
                                     f��(x)
                 -2
                                                                < 0, on (−2, 2).
                                                                                       Concave down.
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